Precalculus · F.TF.B.7

Solving trig equations by factoring, quadratics and identities

An equation holds two trig terms, or a square, or a double angle, so no single sine, cosine or tangent can be got alone. How do you break it into basic equations without losing a solution or picking up a false one?

You need

Double angle formulasPrecalculus The quadratic formulaGrades 9 to 10 Solving quadratics by factoringAlgebra 1

Where you are 4 in Solving trigonometric equations

Trigonometry7

Trigonometric identities13

Solving trigonometric equations4

Solving basic trig equations Trigonometric inequalities Solving equations with a multiple angle

Solving trig equations by factoring, quadratics and identities

When you reach for it

A basic trig equation has one sine, cosine or tangent of x, and algebra gets it alone. Many equations will not allow that. In 2\sin^2 x = \sin x the sine appears twice, once squared. In 2\cos^2 x + 3\sin x = 3 two different functions appear. In \cos 2x + \cos x = 0 the angles differ.

The move for all of them is the one that solves polynomial equations: get 0 on one side, factor, and use the Solving quadratics by factoring. A product is zero only when one of its factors is zero, so one hard equation becomes several basic ones, each finished as before. When the equation is not yet in a form that factors, an identity first rewrites it in one function of one angle.

Part 1 (the first day) factors equations that are already in one or two functions of x, including those that are quadratic in \sin x or \cos x. Part 2 (the second day) uses a Pythagorean or double angle identity first, and ends with squaring both sides, a last resort that needs every answer checked.

The procedure

Part 1: zero on one side, then factor

The zero product property only speaks about a product equal to 0. So every term goes to one side first. Then the expression is factored just as a polynomial would be, treating \sin x and \cos x as unknown numbers: a common factor comes out by the distributive property, four terms may factor by grouping, and a quadratic factors into two binomials. Each factor set equal to 0 is a basic equation.

Part 1: never divide by an expression that can be zero

Dividing both sides by \sin x looks like a shortcut, but it is only allowed when \sin x \neq 0. At the angles where \sin x = 0 the division is a division by zero, and the solutions there vanish from the work without a trace. Factoring keeps that case as a factor of its own, so it is never lost. Example 1 shows the difference.

Part 1: equations that are quadratic in one function

An equation such as 2\cos^2 x + 3\cos x - 2 = 0 is a quadratic in the number u = \cos x. Solve 2u^2 + 3u - 2 = 0 by factoring, or by the The quadratic formula when it does not factor, then return to \cos x = u for each root. A root outside [-1, 1] is rejected: no angle has a sine or cosine of -2 or 3.4. Tangent takes every real value, so for \tan x no root is rejected.

Part 2: one function first, by an identity

When two functions or two angles appear, an identity can rewrite one of them, and the choice is made to leave a single function of a single angle:

the Pythagorean identity trades \cos^2 x = 1 - \sin^2 x, or \sin^2 x = 1 - \cos^2 x, or \sec^2 x = 1 + \tan^2 x;
the double angle formulas trade \sin 2x = 2\sin x\cos x, and \cos 2x for whichever of its three forms matches the rest: 1 - 2\sin^2 x beside a sine, 2\cos^2 x - 1 beside a cosine.

These are identities, true for every x, so the new equation has exactly the same solutions as the old one.

Part 2: squaring, as a last resort

Some equations, such as \displaystyle 1 + \sin x = \sqrt{3}\cos x, mix sine and cosine to the first power, and no identity removes either. Squaring both sides turns \cos x into \cos^2 x = 1 - \sin^2 x. But squaring is not reversible: a^2 = b^2 holds when a = b and also when a = -b. The squared equation has every solution of the original and possibly more, so every candidate is substituted back into the original equation, and those that fail are thrown out.

All answers below are given on [0, 2\pi) or as every solution, with k any integer.

Worked examples

Examples 1 to 4 are Part 1 (factoring and quadratic form). Examples 5 to 7 are Part 2 (an identity first, then squaring).

Example 1

Part 1: a common factor, and what dividing loses

Solve 2\sin^2 x = \sin x on [0, 2\pi), then give all solutions.

Solution

Subtract \sin x from both sides so that one side is 0, then take out the common factor \sin x (distributive property):

\sin x\,(2\sin x - 1) = 0

By the zero product property, \sin x = 0 or 2\sin x - 1 = 0. The first gives x = 0 and x = \pi. The second says \displaystyle \sin x = \frac{1}{2}, with reference angle \displaystyle \frac{\pi}{6} in Quadrants I and II: \displaystyle x = \frac{\pi}{6} and \displaystyle x = \frac{5\pi}{6}. On [0, 2\pi):

x = 0, \ \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \pi

All solutions: x = \pi k, or \displaystyle x = \frac{\pi}{6} + 2\pi k, or \displaystyle x = \frac{5\pi}{6} + 2\pi k. The zeros of sine are \pi apart, so one family with step \pi covers them.

Dividing both sides by \sin x would have given 2\sin x = 1 at once, and only the two solutions \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}. The other two are exactly where \sin x = 0, where that division is not allowed.

Figure 1Example 1. The curves y = 2\sin^2 x and y = \sin x meet four times in one turn. Dividing by \sin x finds only the two crossings at height \displaystyle \frac{1}{2} and loses the two on the axis, where \sin x = 0.

Example 2

Part 1: factoring by grouping

Find all solutions of the equation below.

2\sin x\cos x + \sqrt{2}\sin x - 2\cos x - \sqrt{2} = 0

Solution

Four terms with no factor common to all suggest grouping. The first two share \sin x and the last two share -1:

\sin x\,(2\cos x + \sqrt{2}) - 1\cdot(2\cos x + \sqrt{2}) = 0

Now \displaystyle 2\cos x + \sqrt{2} is a common factor, and the equation is \displaystyle (\sin x - 1)(2\cos x + \sqrt{2}) = 0. Expanding gives back the four terms, which checks the factoring.

The factor \sin x - 1 = 0 gives \sin x = 1, which happens only at the top of the circle, \displaystyle x = \frac{\pi}{2} + 2\pi k. The factor \displaystyle 2\cos x + \sqrt{2} = 0 gives \displaystyle \cos x = -\frac{\sqrt{2}}{2}, with reference angle \displaystyle \frac{\pi}{4} in Quadrants II and III: \displaystyle x = \frac{3\pi}{4} and \displaystyle \frac{5\pi}{4}. Since \displaystyle \frac{5\pi}{4} = -\frac{3\pi}{4} + 2\pi, all solutions are

\begin{aligned} x &= \frac{\pi}{2} + 2\pi k \\ \text{or} \quad x &= \pm\frac{3\pi}{4} + 2\pi k \end{aligned}

Example 3

Part 1: a quadratic in cosine with a root to reject

Solve 2\cos^2 x + 3\cos x - 2 = 0 on [0, 2\pi).

Solution

Write u = \cos x. Then 2u^2 + 3u - 2 = 0, and the factors of 2 \cdot (-2) = -4 that add to 3 are 4 and -1, so 2u^2 + 3u - 2 = (2u - 1)(u + 2). Returning to cosine:

(2\cos x - 1)(\cos x + 2) = 0

The first factor gives \displaystyle \cos x = \frac{1}{2}, so \displaystyle x = \frac{\pi}{3} or \displaystyle x = \frac{5\pi}{3}. The second gives \cos x = -2. Cosine is the horizontal coordinate of a point on the unit circle, which never goes below -1, so this factor has no solutions.

Figure 2Example 3. The factor 2\cos x - 1 gives the line \displaystyle \cos x = \frac{1}{2}, which meets the circle twice. The factor \cos x + 2 gives \cos x = -2, a line the circle never reaches, so it adds nothing.

On [0, 2\pi) the solutions are \displaystyle x = \frac{\pi}{3} and \displaystyle x = \frac{5\pi}{3}.

Example 4

Part 1: the quadratic formula, and a decimal answer

Solve \sin^2 x - 4\sin x + 2 = 0. Give all solutions to four decimal places.

Solution

As a quadratic in u = \sin x, u^2 - 4u + 2 = 0 has discriminant 16 - 8 = 8, not a perfect square, so it does not factor over the integers. The quadratic formula gives

\sin x = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}

Now \displaystyle 2 + \sqrt{2} \approx 3.414 is above 1 and is rejected. The other root, \displaystyle 2 - \sqrt{2} \approx 0.5858, lies in [-1, 1]. With the inverse sine, \displaystyle \alpha = \arcsin(2 - \sqrt{2}) \approx 0.6259, and the mirror angle is \pi - \alpha \approx 2.5157. All solutions:

\begin{aligned} x &\approx 0.6259 + 2\pi k \\ \text{or} \quad x &\approx 2.5157 + 2\pi k \end{aligned}

Check. \sin 0.6259 \approx 0.5858, and 0.5858^2 - 4(0.5858) + 2 is about 0.3431 - 2.3431 + 2 = 0. Rounding happens only at the end; \pi - \alpha is computed from the unrounded \alpha.

Example 5

Part 2: the Pythagorean identity makes one function

Solve 2\cos^2 x + 3\sin x = 3 on [0, 2\pi).

Solution

Two functions appear, but the cosine is squared, and \cos^2 x = 1 - \sin^2 x for every x. Substituting leaves only sines:

2 - 2\sin^2 x + 3\sin x = 3

Move every term to the right side, where \sin^2 x keeps a positive coefficient: 0 = 2\sin^2 x - 3\sin x + 1. This factors as

(2\sin x - 1)(\sin x - 1) = 0

So \displaystyle \sin x = \frac{1}{2}, giving \displaystyle x = \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}, or \sin x = 1, giving only \displaystyle x = \frac{\pi}{2}, since the line y = 1 touches the circle at one point. On [0, 2\pi): \displaystyle x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, three solutions.

Check at \displaystyle \frac{\pi}{2}. \displaystyle 2\cos^2\frac{\pi}{2} + 3\sin\frac{\pi}{2} = 0 + 3 = 3.

Example 6

Part 2: double angles, and choosing the form

Solve on [0, 2\pi): (a) \displaystyle \sin 2x = \sqrt{2}\sin x; (b) \cos 2x + \cos x = 0.

Solution

(a) The double angle formula \sin 2x = 2\sin x\cos x puts every term at angle x. Then subtract and factor out \sin x, rather than divide by it:

\begin{aligned} 2\sin x\cos x - \sqrt{2}\sin x &= 0 \\ \sin x\,(2\cos x - \sqrt{2}) &= 0 \end{aligned}

The factor \sin x = 0 gives x = 0, \pi. The factor \displaystyle \cos x = \frac{\sqrt{2}}{2} gives \displaystyle x = \frac{\pi}{4}, \frac{7\pi}{4}. Four solutions: \displaystyle x = 0, \frac{\pi}{4}, \pi, \frac{7\pi}{4}.

(b) Of the three forms of \cos 2x, the one written in cosines alone, 2\cos^2 x - 1, leaves a single function:

\begin{aligned} 2\cos^2 x + \cos x - 1 &= 0 \\ (2\cos x - 1)(\cos x + 1) &= 0 \end{aligned}

So \displaystyle \cos x = \frac{1}{2}, giving \displaystyle x = \frac{\pi}{3} and \displaystyle \frac{5\pi}{3}, or \cos x = -1, giving x = \pi. On [0, 2\pi): \displaystyle x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}. These are evenly spaced, so all solutions form one family, \displaystyle x = \frac{\pi}{3} + \frac{2\pi}{3}k. Using 1 - 2\sin^2 x instead would have left both \sin^2 x and \cos x in the equation.

Example 7

Part 2: squaring, with every candidate checked

Solve \displaystyle \sec x + \tan x = \sqrt{3} on [0, 2\pi).

Solution

In sines and cosines, \displaystyle \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sqrt{3}. Both terms exist only when \cos x \neq 0, so \displaystyle x = \frac{\pi}{2} and \displaystyle \frac{3\pi}{2} are excluded from the start. Multiplying by \cos x (allowed, since it is not 0):

1 + \sin x = \sqrt{3}\cos x

Sine and cosine both appear to the first power. Squaring both sides turns the cosine into a square that the Pythagorean identity can trade:

\begin{aligned} 1 + 2\sin x + \sin^2 x &= 3\cos^2 x \\ &= 3 - 3\sin^2 x \end{aligned}

Collect terms: 4\sin^2 x + 2\sin x - 2 = 0, and dividing by the number 2 (never 0) gives 2\sin^2 x + \sin x - 1 = 0, which factors as (2\sin x - 1)(\sin x + 1) = 0. The candidates are \displaystyle \sin x = \frac{1}{2}, at \displaystyle x = \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}, and \sin x = -1, at \displaystyle x = \frac{3\pi}{2}.

Squaring may have added solutions, so each candidate goes back into the original equation.

Candidate	\sec x + \tan x	Keep?
\displaystyle \frac{\pi}{6}	\displaystyle \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \sqrt{3}	yes
\displaystyle \frac{5\pi}{6}	\displaystyle -\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = -\sqrt{3}	no
\displaystyle \frac{3\pi}{2}	undefined	no

The only solution on [0, 2\pi) is \displaystyle x = \frac{\pi}{6}, and all solutions are \displaystyle x = \frac{\pi}{6} + 2\pi k. The false candidate \displaystyle \frac{5\pi}{6} solves \displaystyle 1 + \sin x = -\sqrt{3}\cos x, the equation with the other sign, which squares to the same thing.

Boundaries

Zero on one side. The zero product property needs a product equal to 0. From \sin x\,(\sin x + 1) = 2 nothing follows about either factor alone.
Dividing. Dividing by \sin x, \cos x or any expression that can be zero is allowed only on the angles where it is not zero, and those angles must then be tested separately. Dividing by a nonzero number, such as 2, is always safe.
The range. A root of a quadratic in \sin x or \cos x that lies outside [-1, 1] gives no angle. A root for \tan x is never rejected for its size.
Where each function exists. \tan x and \sec x need \cos x \neq 0; \cot x and \csc x need \sin x \neq 0. A solution that makes a term of the original equation undefined is not a solution, even when it satisfies a rewritten form.
Squaring. The squared equation may have extra solutions, those of the equation with one side’s sign reversed. Only checking in the original equation tells them apart. Factoring and identities never add or lose solutions, provided no division by zero occurs.
Not every equation factors. An equation such as \displaystyle \cos x = \frac{x}{2} mixes x inside and outside the function, and no factoring or identity solves it exactly; it is solved numerically.

Where this goes next

A trigonometric inequality such as \cos 2x + \cos x > 0 starts from the equation solved here: its solutions cut the turn into stretches, and a sign chart picks the stretches that work. An equation like Example 7, with sine and cosine both to the first power, can also be solved without squaring, by writing \displaystyle \sqrt{3}\cos x - \sin x as one wave; Combining a sine and a cosine into one wave shows how, and that route creates no false candidates. In calculus, finding where a function built from sines and cosines levels off leads to exactly these equations.

Exercises

Part 1: factoring

Solve \displaystyle \sqrt{3}\tan^2 x = \tan x on [0, 2\pi).
Find all solutions of 4\sin x\cos x + 2\cos x = 0.
Solve \sin x\tan x = \sin x on [0, 2\pi).
Solve \displaystyle 2\sin x\cos x - \sqrt{3}\sin x + 2\cos x - \sqrt{3} = 0 on [0, 2\pi).
Find every \theta with 0^\circ \le \theta < 360^\circ such that \displaystyle 2\cos^2\theta + \sqrt{2}\cos\theta = 0.

Part 1: quadratic in one function

Give decimal answers to four decimal places.

Solve 2\cos^2 x - \cos x - 1 = 0 on [0, 2\pi).
Find all solutions of 2\sin^2 x + 7\sin x + 3 = 0.
Find all solutions of 4\cos^2 x - 3 = 0.
Find all solutions of 3\tan^2 x - 4\tan x + 1 = 0.
Solve \cos^2 x + 2\cos x - 1 = 0 on [0, 2\pi).
Solve 3\sin^2 x - \sin x - 1 = 0 on [0, 2\pi).

Part 2: an identity first

Solve \sin^2 x = 1 + \cos x on [0, 2\pi).
Find all solutions of \sec^2 x = 2\tan x.
Solve \cos 2x = 5\cos x + 2 on [0, 2\pi).
Find all solutions of \cos 2x + 5\sin x = 3.
Find all solutions of \sin 2x + \sin x = 0.
Solve \sin 2x = \tan x on [0, 2\pi).
Solve \cos 2x = 2\sin x on [0, 2\pi), to four decimal places.
Solve \cos x - \sin x = 1 on [0, 2\pi) by squaring both sides, and check every candidate.
Find every \theta with 0^\circ \le \theta < 360^\circ such that \sec^2\theta + \tan\theta = 3, to two decimal places where not exact.

Reasoning

Galymzhan solves \displaystyle \sin x\cos x = \frac{1}{2}\cos x on [0, 2\pi) by dividing both sides by \cos x, and reports \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}. Which solutions did he lose, and why? Solve it correctly.
Explain why an equation that is quadratic in \sin x has at most four solutions on [0, 2\pi), and give an example with exactly three.
Elmira squares both sides of \sin x = \cos x and gets \sin^2 x = \cos^2 x, which has four solutions on [0, 2\pi). Which of them solve the original equation, and where did the others come from?
Indira replaces \cos 2x by 1 - 2\sin^2 x in \cos 2x = 4\cos x - 3 and cannot finish. Explain which form of \cos 2x she should use and why, then find all solutions.
Explain why dividing 3\cos^2 x = \sin^2 x by \cos^2 x loses no solution, even though \cos^2 x can be 0. Then find all solutions.

Answers

\displaystyle x = 0, \frac{\pi}{6}, \pi, \frac{7\pi}{6}. \displaystyle \tan x\,(\sqrt{3}\tan x - 1) = 0, so \tan x = 0 or \displaystyle \tan x = \frac{1}{\sqrt{3}}.
\displaystyle x = \frac{\pi}{2} + \pi k, \displaystyle x = \frac{7\pi}{6} + 2\pi k or \displaystyle x = \frac{11\pi}{6} + 2\pi k. 2\cos x\,(2\sin x + 1) = 0, so \cos x = 0 or \displaystyle \sin x = -\frac{1}{2}.
\displaystyle x = 0, \frac{\pi}{4}, \pi, \frac{5\pi}{4}. \sin x\,(\tan x - 1) = 0; tangent exists at all four angles.
\displaystyle x = \frac{\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}. Grouping gives \displaystyle \sin x\,(2\cos x - \sqrt{3}) + (2\cos x - \sqrt{3}) = 0, that is, \displaystyle (\sin x + 1)(2\cos x - \sqrt{3}) = 0.
\theta = 90^\circ, 135^\circ, 225^\circ, 270^\circ. \displaystyle \cos\theta\,(2\cos\theta + \sqrt{2}) = 0, so \cos\theta = 0 or \displaystyle \cos\theta = -\frac{\sqrt{2}}{2}.
\displaystyle x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}. (2\cos x + 1)(\cos x - 1) = 0.
\displaystyle x = \frac{7\pi}{6} + 2\pi k or \displaystyle x = \frac{11\pi}{6} + 2\pi k. (2\sin x + 1)(\sin x + 3) = 0; \sin x = -3 is rejected.
\displaystyle x = \frac{\pi}{6} + \pi k or \displaystyle x = \frac{5\pi}{6} + \pi k. \displaystyle \cos x = \pm\frac{\sqrt{3}}{2} gives \displaystyle \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, which pair off \pi apart.
\displaystyle x = \frac{\pi}{4} + \pi k or x \approx 0.3218 + \pi k. (3\tan x - 1)(\tan x - 1) = 0, and \displaystyle \arctan\frac{1}{3} \approx 0.3218.
x \approx 1.1437 or x \approx 5.1395. \displaystyle \cos x = -1 \pm \sqrt{2}; \displaystyle -1 - \sqrt{2} < -1 is rejected, and \displaystyle \arccos(\sqrt{2} - 1) \approx 1.1437, with 2\pi - 1.1437 \approx 5.1395.
x \approx 0.8751, 2.2665, 3.5908, 5.8340. \displaystyle \sin x = \frac{1 \pm \sqrt{13}}{6} \approx 0.7676 or -0.4343, both in range; \arcsin 0.7676 \approx 0.8751 and its mirror \pi - 0.8751, then \arcsin(-0.4343) \approx -0.4492, giving \pi + 0.4492 and 2\pi - 0.4492.
\displaystyle x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}. 1 - \cos^2 x = 1 + \cos x gives \cos x\,(\cos x + 1) = 0.
\displaystyle x = \frac{\pi}{4} + \pi k. 1 + \tan^2 x = 2\tan x gives (\tan x - 1)^2 = 0, so \tan x = 1.
\displaystyle x = \frac{2\pi}{3}, \frac{4\pi}{3}. 2\cos^2 x - 1 = 5\cos x + 2 gives (2\cos x + 1)(\cos x - 3) = 0; \cos x = 3 is rejected.
\displaystyle x = \frac{\pi}{6} + 2\pi k or \displaystyle x = \frac{5\pi}{6} + 2\pi k. 1 - 2\sin^2 x + 5\sin x = 3 gives (2\sin x - 1)(\sin x - 2) = 0; \sin x = 2 is rejected.
x = \pi k, \displaystyle x = \frac{2\pi}{3} + 2\pi k or \displaystyle x = \frac{4\pi}{3} + 2\pi k. \sin x\,(2\cos x + 1) = 0.
\displaystyle x = 0, \frac{\pi}{4}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{7\pi}{4}. With \cos x \neq 0, multiply by \cos x: 2\sin x\cos^2 x = \sin x, so \sin x\,(2\cos^2 x - 1) = 0, giving \sin x = 0 or \displaystyle \cos x = \pm\frac{\sqrt{2}}{2}. None of these has \cos x = 0.
x \approx 0.3747 or x \approx 2.7669. 1 - 2\sin^2 x = 2\sin x gives 2\sin^2 x + 2\sin x - 1 = 0, so \displaystyle \sin x = \frac{-1 \pm \sqrt{3}}{2}; \displaystyle \frac{-1 - \sqrt{3}}{2} < -1 is rejected, and \displaystyle \arcsin\frac{\sqrt{3} - 1}{2} \approx 0.3747.
\displaystyle x = 0, \frac{3\pi}{2}. Squaring gives 1 - 2\sin x\cos x = 1, so \sin x\cos x = 0 and the candidates are \displaystyle 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}. At \displaystyle \frac{\pi}{2} and \pi the left side is -1, so they are rejected.
\theta = 45^\circ, 225^\circ, \theta \approx 116.57^\circ, 296.57^\circ. 1 + \tan^2\theta + \tan\theta = 3 gives (\tan\theta + 2)(\tan\theta - 1) = 0; \arctan(-2) \approx -63.43^\circ, plus 180^\circ and 360^\circ.
He lost \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}, where \cos x = 0: there the division is by zero, and both sides of the original equation are 0. Correctly, \displaystyle \cos x\,(\sin x - \frac{1}{2}) = 0, so \displaystyle x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}.
A quadratic in u = \sin x has at most two roots u, and each value u is the sine of at most two angles in one turn, so there are at most 2 \cdot 2 = 4 solutions. Exactly three happen when one root is 1 or -1 (one angle) and the other lies strictly between -1 and 1 (two angles), as in \sin^2 x - \sin x = 0, with solutions \displaystyle 0, \frac{\pi}{2}, \pi.
\sin^2 x = \cos^2 x holds at \displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, where \cos 2x = \cos^2 x - \sin^2 x = 0. Only \displaystyle x = \frac{\pi}{4}, \frac{5\pi}{4} solve \sin x = \cos x. The other two solve \sin x = -\cos x, which squares to the same equation.
She should use 2\cos^2 x - 1, because the rest of the equation is in \cos x, so this form leaves a single function. Then 2\cos^2 x - 1 = 4\cos x - 3 becomes 2(\cos x - 1)^2 = 0, so \cos x = 1 and x = 2\pi k.
If \cos x = 0 then \sin^2 x = 1, and the equation would say 0 = 1, so no solution has \cos x = 0 and nothing is lost by dividing. Then \tan^2 x = 3, \displaystyle \tan x = \pm\sqrt{3}, and \displaystyle x = \frac{\pi}{3} + \pi k or \displaystyle x = \frac{2\pi}{3} + \pi k.