Solving Quadratic Trig Equations

Some trig equations contain a squared trig ratio, like $2\cos^2 x - \cos x - 1 = 0$. These are quadratic trig equations, and you solve them the same way you solve any quadratic: factor, then set each factor equal to zero. Each factor gives a linear trig equation, which you already know how to solve. Often you’ll need an identity first to get the equation into a form you can factor. All answers are in radians, for $0 \le x \le 2\pi$.

Key ideas

Treat the trig ratio like a variable

$2\cos^2 x - \cos x - 1 = 0$ has the same shape as $2u^2 - u - 1 = 0$, with $u = \cos x$. Factor it the same way:

$$2u^2 - u - 1 = (2u + 1)(u - 1) \quad\Rightarrow\quad 2\cos^2 x - \cos x - 1 = (2\cos x + 1)(\cos x - 1)$$

Then use the zero product property: if a product is $0$, at least one factor is $0$. Solve each factor as a linear trig equation.

If a factor gives $\sin x$ or $\cos x$ less than $-1$ or greater than $1$, that factor has no solutions: reject it and keep the others.

Common factor; never divide by a trig ratio

In an equation like $2\sin x\cos x = \sqrt{3}\cos x$, it’s tempting to divide both sides by $\cos x$. Don’t. If $\cos x = 0$, you’d be dividing by zero, and you’d lose the solutions where $\cos x = 0$. Instead, move everything to one side and common factor:

$$2\sin x\cos x - \sqrt{3}\cos x = 0 \quad\Rightarrow\quad \cos x\,(2\sin x - \sqrt{3}) = 0$$

Use an identity to get one trig ratio

If an equation mixes ratios or angles, use an identity to rewrite it in terms of a single ratio of a single angle:

If the equation has…	try…
$\sin^2 x$ and $\cos x$	$\sin^2 x = 1 - \cos^2 x$
$\cos^2 x$ and $\sin x$	$\cos^2 x = 1 - \sin^2 x$
$\cos 2x$ and $\sin x$	$\cos 2x = 1 - 2\sin^2 x$
$\cos 2x$ and $\cos x$	$\cos 2x = 2\cos^2 x - 1$
$\sin 2x$ and $\sin x$ or $\cos x$	$\sin 2x = 2\sin x\cos x$, then common factor

Two ways to handle cos 2x on its own

An equation like $\cos 2x = -\dfrac{1}{2}$ is really linear in $2x$, so you can solve it with $u = 2x$ on $0 \le u \le 4\pi$, as in linear trig equations. You can also replace $\cos 2x$ with $1 - 2\sin^2 x$ and solve a quadratic. Both methods give the same answers (see Practice question 9).

Steps

1. Use identities, if needed, to get a single ratio of a single angle.
2. Move everything to one side so the other side is $0$.
3. Factor (common factor, trinomial, or difference of squares).
4. Set each factor to $0$, reject impossible values, and solve each linear equation on $0 \le x \le 2\pi$.
5. List all the solutions, and check them in the original equation (or on a graph).

Worked examples

Example 1: Factoring a trinomial

Solve $2\cos^2 x - \cos x - 1 = 0$ for $0 \le x \le 2\pi$.

Solution. Factor as above:

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

First factor: $2\cos x + 1 = 0$, so $\cos x = -\dfrac{1}{2}$. The related angle is $\dfrac{\pi}{3}$, and cosine is negative in quadrants 2 and 3:

$$x = \frac{2\pi}{3} \quad\text{or}\quad x = \frac{4\pi}{3}$$

Second factor: $\cos x - 1 = 0$, so $\cos x = 1$. That’s the point $(1, 0)$ on the unit circle, which is at both ends of the interval:

$$x = 0 \quad\text{or}\quad x = 2\pi$$

Answer: $x = 0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ 2\pi$.

Check $x = \dfrac{2\pi}{3}$: $2\left(-\dfrac{1}{2}\right)^2 - \left(-\dfrac{1}{2}\right) - 1 = \dfrac{1}{2} + \dfrac{1}{2} - 1 = 0$. ✓

Example 2: Common factoring

Solve $2\sin x\cos x = \sqrt{3}\cos x$ for $0 \le x \le 2\pi$.

Solution. Move everything to one side and common factor (don’t divide by $\cos x$):

$$2\sin x\cos x - \sqrt{3}\cos x = 0 \quad\Rightarrow\quad \cos x\,(2\sin x - \sqrt{3}) = 0$$

First factor: $\cos x = 0$ at the top and bottom of the unit circle:

$$x = \frac{\pi}{2} \quad\text{or}\quad x = \frac{3\pi}{2}$$

Second factor: $\sin x = \dfrac{\sqrt{3}}{2}$, with related angle $\dfrac{\pi}{3}$ in quadrants 1 and 2:

$$x = \frac{\pi}{3} \quad\text{or}\quad x = \frac{2\pi}{3}$$

Answer: $x = \dfrac{\pi}{3},\ \dfrac{\pi}{2},\ \dfrac{2\pi}{3},\ \dfrac{3\pi}{2}$.

If you had divided by $\cos x$ at the start, you’d have found only $\dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$, and missed half the answers.

Example 3: Using a Pythagorean identity

Solve $2\sin^2 x = 3\cos x$ for $0 \le x \le 2\pi$.

Solution. The equation mixes $\sin^2 x$ and $\cos x$. Replace $\sin^2 x$ with $1 - \cos^2 x$:

$$\begin{aligned} 2(1 - \cos^2 x) &= 3\cos x \\ 2 - 2\cos^2 x - 3\cos x &= 0 \\ 2\cos^2 x + 3\cos x - 2 &= 0 && \text{multiply by } -1 \\ (2\cos x - 1)(\cos x + 2) &= 0 \end{aligned}$$

First factor: $\cos x = \dfrac{1}{2}$, with related angle $\dfrac{\pi}{3}$ in quadrants 1 and 4:

$$x = \frac{\pi}{3} \quad\text{or}\quad x = \frac{5\pi}{3}$$

Second factor: $\cos x = -2$. This is impossible, since $\cos x$ is never less than $-1$. Reject it.

Answer: $x = \dfrac{\pi}{3},\ \dfrac{5\pi}{3}$.

$y = 2\sin^2 x$ and $y = 3\cos x$ meet only at $x = \dfrac{\pi}{3}$ and $x = \dfrac{5\pi}{3}$.

Check $x = \dfrac{\pi}{3}$: $2\left(\dfrac{\sqrt{3}}{2}\right)^2 = 2\cdot\dfrac{3}{4} = 1.5$ and $3\cos\dfrac{\pi}{3} = 3\cdot\dfrac{1}{2} = 1.5$. ✓

Example 4: Using a double angle identity

Solve $\cos 2x + 3\sin x - 2 = 0$ for $0 \le x \le 2\pi$.

Solution. The other term is $\sin x$, so use $\cos 2x = 1 - 2\sin^2 x$:

$$\begin{aligned} 1 - 2\sin^2 x + 3\sin x - 2 &= 0 \\ -2\sin^2 x + 3\sin x - 1 &= 0 \\ 2\sin^2 x - 3\sin x + 1 &= 0 && \text{multiply by } -1 \\ (2\sin x - 1)(\sin x - 1) &= 0 \end{aligned}$$

First factor: $\sin x = \dfrac{1}{2}$, so $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$.

Second factor: $\sin x = 1$, so $x = \dfrac{\pi}{2}$.

Answer: $x = \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6}$.

Check $x = \dfrac{\pi}{2}$ in the original equation: $\cos\pi + 3\sin\dfrac{\pi}{2} - 2 = -1 + 3 - 2 = 0$. ✓

Common mistakes

Dividing by a trig ratio. Dividing both sides by $\cos x$ (or $\sin x$) throws away the solutions where it equals zero. Move everything to one side and common factor instead.

Keeping impossible values. $\cos x = -2$ or $\sin x = \dfrac{3}{2}$ have no solutions. Reject those factors, but don’t forget to solve the others.

Mixing ratios or angles when factoring. An equation with both $\cos 2x$ and $\sin x$ can’t be factored as it stands. Use an identity first so everything is in terms of one ratio of one angle.

Choosing the wrong form of cos 2x. Pick the form that matches the other terms: $1 - 2\sin^2 x$ if the rest is in sines, $2\cos^2 x - 1$ if the rest is in cosines.

Taking only the positive square root. $\sin^2 x = \dfrac{1}{4}$ means $\sin x = \dfrac{1}{2}$ or $\sin x = -\dfrac{1}{2}$. Missing the negative root loses half the solutions.

Not listing every solution. Each factor usually gives two angles, and values like $\cos x = 1$ include both endpoints. Count your solutions and compare with a graph.

Practice

1. (Warm-up) Solve $(\sin x - 1)(2\cos x + 1) = 0$ for $0 \le x \le 2\pi$.

Solution

$\sin x = 1$ gives $x = \dfrac{\pi}{2}$.

$\cos x = -\dfrac{1}{2}$ gives $x = \dfrac{2\pi}{3}$ or $x = \dfrac{4\pi}{3}$ (quadrants 2 and 3).

So $x = \dfrac{\pi}{2},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}$.

2. (Warm-up) Solve $4\sin^2 x - 1 = 0$ for $0 \le x \le 2\pi$.

Solution

$\sin^2 x = \dfrac{1}{4}$, so $\sin x = \pm\dfrac{1}{2}$. (You can also factor: $(2\sin x - 1)(2\sin x + 1) = 0$.)

$\sin x = \dfrac{1}{2}$: $x = \dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$.

$\sin x = -\dfrac{1}{2}$: $x = \dfrac{7\pi}{6}$ or $\dfrac{11\pi}{6}$.

So $x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}$.

3. (Core) Solve $2\sin^2 x - \sin x = 0$ for $0 \le x \le 2\pi$.

Solution

Common factor: $\sin x\,(2\sin x - 1) = 0$.

$\sin x = 0$: $x = 0$, $\pi$, $2\pi$.

$\sin x = \dfrac{1}{2}$: $x = \dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$.

So $x = 0,\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \pi,\ 2\pi$.

4. (Core) Solve $2\sin^2 x - 3\sin x - 2 = 0$ for $0 \le x \le 2\pi$.

Solution

Factor: $(2\sin x + 1)(\sin x - 2) = 0$.

$\sin x = 2$ is impossible, so reject it.

$\sin x = -\dfrac{1}{2}$: related angle $\dfrac{\pi}{6}$, quadrants 3 and 4, so $x = \dfrac{7\pi}{6}$ or $\dfrac{11\pi}{6}$.

Check $x = \dfrac{7\pi}{6}$: $2\left(\dfrac{1}{4}\right) - 3\left(-\dfrac{1}{2}\right) - 2 = \dfrac{1}{2} + \dfrac{3}{2} - 2 = 0$. ✓

5. (Core) Solve $\cos 2x + \cos x = 0$ for $0 \le x \le 2\pi$.

Solution

The other term is $\cos x$, so use $\cos 2x = 2\cos^2 x - 1$:

$$2\cos^2 x + \cos x - 1 = 0 \quad\Rightarrow\quad (2\cos x - 1)(\cos x + 1) = 0$$

$\cos x = \dfrac{1}{2}$: $x = \dfrac{\pi}{3}$ or $\dfrac{5\pi}{3}$.

$\cos x = -1$: $x = \pi$.

So $x = \dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{3}$.

Check $x = \pi$: $\cos 2\pi + \cos\pi = 1 - 1 = 0$. ✓

6. (Core) Solve $\sin 2x + \cos x = 0$ for $0 \le x \le 2\pi$.

Solution

Use $\sin 2x = 2\sin x\cos x$, then common factor:

$$2\sin x\cos x + \cos x = 0 \quad\Rightarrow\quad \cos x\,(2\sin x + 1) = 0$$

$\cos x = 0$: $x = \dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$.

$\sin x = -\dfrac{1}{2}$: $x = \dfrac{7\pi}{6}$ or $\dfrac{11\pi}{6}$.

So $x = \dfrac{\pi}{2},\ \dfrac{7\pi}{6},\ \dfrac{3\pi}{2},\ \dfrac{11\pi}{6}$.

7. (Core) Solve $5\sin^2 x + 3\sin x - 2 = 0$ for $0 \le x \le 2\pi$. Give exact answers where possible, and otherwise round to three decimal places.

Solution

Factor: $(5\sin x - 2)(\sin x + 1) = 0$.

$\sin x = \dfrac{2}{5} = 0.4$: the related angle is $\sin^{-1}(0.4) \approx 0.4115$ (radian mode). Sine is positive in quadrants 1 and 2:

$$x \approx 0.412 \qquad\text{or}\qquad x \approx \pi - 0.4115 \approx 2.730$$

$\sin x = -1$: $x = \dfrac{3\pi}{2}$ (about $4.712$).

So $x \approx 0.412$, $x \approx 2.730$, or $x = \dfrac{3\pi}{2}$.

8. (Core) When light passes through two polarizing filters (like the lenses of some sunglasses), the intensity that gets through is $I = I_0\cos^2\theta$, where $I_0$ is the intensity after the first filter and $\theta$ is the angle between the filters. For which angles $\theta$ with $0 \le \theta \le 2\pi$ does exactly half of $I_0$ get through?

Solution

Set $I = \dfrac{1}{2}I_0$:

$$I_0\cos^2\theta = \frac{1}{2}I_0 \quad\Rightarrow\quad \cos^2\theta = \frac{1}{2} \quad\Rightarrow\quad \cos\theta = \pm\frac{\sqrt{2}}{2}$$

(Dividing by $I_0$ is fine: it’s a positive constant, not a trig expression that could be $0$.)

$\cos\theta = \dfrac{\sqrt{2}}{2}$: $\theta = \dfrac{\pi}{4}$ or $\dfrac{7\pi}{4}$.

$\cos\theta = -\dfrac{\sqrt{2}}{2}$: $\theta = \dfrac{3\pi}{4}$ or $\dfrac{5\pi}{4}$.

So $\theta = \dfrac{\pi}{4},\ \dfrac{3\pi}{4},\ \dfrac{5\pi}{4},\ \dfrac{7\pi}{4}$: whenever the filters are at a $45^\circ$ angle to each other, in either direction.

9. (Challenge) Solve $\cos 2x = -\dfrac{1}{2}$ for $0 \le x \le 2\pi$ in two ways:

• (a) by letting $u = 2x$;
• (b) by using $\cos 2x = 1 - 2\sin^2 x$.

Solution

(a) Let $u = 2x$, with $0 \le u \le 4\pi$. $\cos u = -\dfrac{1}{2}$ in quadrants 2 and 3 with related angle $\dfrac{\pi}{3}$: $u = \dfrac{2\pi}{3}$, $\dfrac{4\pi}{3}$, and adding $2\pi$, $\dfrac{8\pi}{3}$, $\dfrac{10\pi}{3}$. Divide by $2$:

$$x = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{5\pi}{3}$$

(b)

$$1 - 2\sin^2 x = -\frac{1}{2} \quad\Rightarrow\quad -2\sin^2 x = -\frac{3}{2} \quad\Rightarrow\quad \sin^2 x = \frac{3}{4} \quad\Rightarrow\quad \sin x = \pm\frac{\sqrt{3}}{2}$$

$\sin x = \dfrac{\sqrt{3}}{2}$: $x = \dfrac{\pi}{3}$ or $\dfrac{2\pi}{3}$. $\quad\sin x = -\dfrac{\sqrt{3}}{2}$: $x = \dfrac{4\pi}{3}$ or $\dfrac{5\pi}{3}$.

Both methods give the same four solutions. ✓

10. (Challenge) Solve $\cos 2x = 3\sin x + 2$ for $0 \le x \le 2\pi$.

Solution

Use $\cos 2x = 1 - 2\sin^2 x$ and move everything to one side:

$$\begin{aligned} 1 - 2\sin^2 x &= 3\sin x + 2 \\ 0 &= 2\sin^2 x + 3\sin x + 1 \\ 0 &= (2\sin x + 1)(\sin x + 1) \end{aligned}$$

$\sin x = -\dfrac{1}{2}$: $x = \dfrac{7\pi}{6}$ or $\dfrac{11\pi}{6}$.

$\sin x = -1$: $x = \dfrac{3\pi}{2}$.

So $x = \dfrac{7\pi}{6},\ \dfrac{3\pi}{2},\ \dfrac{11\pi}{6}$.

Check $x = \dfrac{3\pi}{2}$: $\cos 3\pi = -1$ and $3(-1) + 2 = -1$. ✓