Math Core

Some equations mix different functions or different angles, like $\sin 2x = \cos x$ or $2\sin^2 x + 3\cos x - 3 = 0$ . You can't isolate "the" trig function because there are two of them. The identities from the previous unit fix this: they let you rewrite the equation in terms of one function of one angle, and then the methods you already know take over.

The goal: one function, one angle

Every strategy in this lesson aims at the same target. Rewrite the equation so it contains a single trig function of a single angle, then either isolate it or factor.

Tools for rewriting trig equations

Pythagorean identities trade squares: $\sin^2 x = 1 - \cos^2 x$ , $\cos^2 x = 1 - \sin^2 x$ , $\tan^2 x = \sec^2 x - 1$ .
Double-angle identities turn $2x$ into $x$ : $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1 = \cos^2 x - \sin^2 x$ .
Get zero on one side and factor. Set each factor equal to $0$ .
Never divide by a trig expression that could be zero. Factor it out instead.

Using a Pythagorean identity

When an equation has $\sin^2 x$ and $\cos x$ (or $\cos^2 x$ and $\sin x$ ), replace the square so that everything is in terms of the other function. The result is a quadratic in one trig function.

Worked example: Trading sin² x for cosine

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

Replace $\sin^2 x$ with $1 - \cos^2 x$ :

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

$\cos x = \dfrac{1}{2}$ : $x = \dfrac{\pi}{3}$ or $x = \dfrac{5\pi}{3}$
$\cos x = 1$ : $x = 0$

The solutions are $0$ , $\dfrac{\pi}{3}$ and $\dfrac{5\pi}{3}$ .

Using a double-angle identity

When the angles don't match, like $2x$ and $x$ , a double-angle identity brings everything down to $x$ .

Worked example: Matching the angles

Solve $\sin 2x = \cos x$ on $[0, 2\pi)$ .

Replace $\sin 2x$ with $2\sin x\cos x$ , move everything to one side and factor:

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

$\cos x = 0$ : $x = \dfrac{\pi}{2}$ or $x = \dfrac{3\pi}{2}$
$\sin x = \dfrac{1}{2}$ : $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$

There are four solutions: $\dfrac{\pi}{6}$ , $\dfrac{\pi}{2}$ , $\dfrac{5\pi}{6}$ and $\dfrac{3\pi}{2}$ . The graph confirms four crossings.

y = sin 2x and y = cos x cross four times on [0, 2π): at π/6, π/2, 5π/6 and 3π/2.Open in grapher →

Common mistake

It's tempting to divide both sides of $2\sin x\cos x = \cos x$ by $\cos x$ to get $2\sin x = 1$ . That throws away $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$ , where $\cos x = 0$ . Dividing by zero is never allowed, and you can't know in advance that $\cos x \ne 0$ . Factor instead of dividing.

Choosing the right form of cos 2x

$\cos 2x$ has three forms. Pick the one that matches the other function in the equation.

If the equation also has $\sin x$ , use $\cos 2x = 1 - 2\sin^2 x$ .
If it also has $\cos x$ , use $\cos 2x = 2\cos^2 x - 1$ .

Worked example: Picking a form of cos 2x

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

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

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

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

The solutions are $\dfrac{\pi}{2}$ , $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$ .

Squaring both sides

Sometimes squaring is the only easy way to connect $\sin x$ and $\cos x$ . But squaring can create extraneous solutions, just like it does with radical equations: $a = b$ and $a = -b$ both square to $a^2 = b^2$ . So after squaring, you must check every answer in the original equation.

Worked example: Checking for extraneous solutions

Solve $\sin x + \cos x = 1$ on $[0, 2\pi)$ .

Square both sides and use $\sin^2 x + \cos^2 x = 1$ :

\begin{aligned} \sin^2 x + 2\sin x\cos x + \cos^2 x &= 1 \\ 1 + 2\sin x\cos x &= 1 \\ \sin 2x &= 0 \end{aligned}

With $u = 2x$ and $0 \le u < 4\pi$ : $u = 0, \pi, 2\pi, 3\pi$ , so $x = 0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}$ .

Now check each one in $\sin x + \cos x = 1$ :

$x$	$\sin x + \cos x$	works?
$0$	$0 + 1 = 1$	yes
$\dfrac{\pi}{2}$	$1 + 0 = 1$	yes
$\pi$	$0 + (-1) = -1$	no
$\dfrac{3\pi}{2}$	$-1 + 0 = -1$	no

The solutions are $x = 0$ and $x = \dfrac{\pi}{2}$ . The other two solve $\sin x + \cos x = -1$ instead.

Tip

A sum formula can avoid squaring. Since $\sin x + \cos x = \sqrt{2}\sin\left(x + \dfrac{\pi}{4}\right)$ , the equation above becomes $\sin\left(x + \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$ , a shifted-angle equation with no extraneous roots.