Equations that use identities

Example 1: Find the solutions of ${2\cos^2 x+\sin x=1}$ in ${[0,2\pi)}$.

Solution:

The equation has both ${\cos x}$ and ${\sin x}$. Replace ${\cos^2 x}$ by ${1-\sin^2 x}$ so that only sine is left:

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

Multiply by ${-1}$, and factor:

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

So ${\sin x=-\dfrac{1}{2}}$ or ${\sin x=1}$. As in the last lesson, the solutions are

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

Example 2: Find the solutions of ${\sec^2 x-2\tan x=4}$ in ${[0,2\pi)}$. Round to four decimal places where needed.

Solution:

Use ${\sec^2 x=1+\tan^2 x}$ to get only tangent:

$\begin{align*}1+\tan^2 x-2\tan x&=4\\\tan^2 x-2\tan x-3&=0\\(\tan x-3)(\tan x+1)&=0\end{align*}$

${\tan x=-1}$: ${x=\dfrac{3\pi}{4}}$ or ${x=\dfrac{7\pi}{4}}$.

${\tan x=3}$: In radian mode, ${\tan^{-1}3\approx 1.2490}$. Tangent has period ${\pi}$, so the other solution in ${[0,2\pi)}$ is ${1.2490+\pi\approx 4.3906}$.

The solutions are ${x\approx 1.2490}$, ${\dfrac{3\pi}{4}}$, ${4.3906}$, ${\dfrac{7\pi}{4}}$.

Example 3: Find the solutions of ${\sin 2x=\cos x}$ in ${[0,2\pi)}$.

Solution:

Use ${\sin 2x=2\sin x\cos x}$, move everything to one side, and factor out ${\cos x}$:

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

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

The solutions are ${\dfrac{\pi}{6}}$, ${\dfrac{\pi}{2}}$, ${\dfrac{5\pi}{6}}$, ${\dfrac{3\pi}{2}}$.

Example 4: Find the solutions of ${\cos 2x+\cos x=0}$ in ${[0,2\pi)}$.

Solution:

Choose the form ${\cos 2x=2\cos^2 x-1}$, so that only cosine appears:

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

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

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

Example 5: Find the solutions of ${\sin x+\cos x=1}$ in ${[0,2\pi)}$.

Solution:

Square both sides: ${(\sin x+\cos x)^2=1}$. Expand the left side:

$\begin{align*}&(\sin x+\cos x)^2\\&=\sin^2 x+2\sin x\cos x+\cos^2 x\\&=1+2\sin x\cos x\end{align*}$

So the squared equation is

$\begin{align*}1+2\sin x\cos x&=1\\\sin x\cos x&=0\end{align*}$

So ${\sin x=0}$ or ${\cos x=0}$. That gives four candidates: ${0}$, ${\dfrac{\pi}{2}}$, ${\pi}$, ${\dfrac{3\pi}{2}}$.

Check each one in ${\sin x+\cos x=1}$:

• ${x=0}$: ${0+1=1}$. Yes.
• ${x=\dfrac{\pi}{2}}$: ${1+0=1}$. Yes.
• ${x=\pi}$: ${0+(-1)=-1}$. No.
• ${x=\dfrac{3\pi}{2}}$: ${-1+0=-1}$. No.

The solutions are ${0}$ and ${\dfrac{\pi}{2}}$. The other two are extraneous.

Example 6: In Double-angle and half-angle formulas, a ball thrown at ${64}$ ft/s at angle ${\theta}$ lands ${R=128\sin 2\theta}$ feet away. At what angles should it be thrown to land ${100}$ ft away?

Solution:

Solve ${128\sin 2\theta=100}$ for ${\theta}$ between ${0^\circ}$ and ${90^\circ}$:

${\sin 2\theta=\dfrac{100}{128}=0.78125}$

Then ${2\theta}$ is between ${0^\circ}$ and ${180^\circ}$. In degree mode, ${\sin^{-1}(0.78125)\approx 51.38^\circ}$. Sine is also positive in quadrant II, so the second solution is ${180^\circ-51.38^\circ=128.62^\circ}$:

${2\theta\approx 51.38^\circ}$or${2\theta\approx 128.62^\circ}$

Divide by ${2}$: ${\theta\approx 25.7^\circ}$ or ${\theta\approx 64.3^\circ}$.

Both angles work: a low, fast throw and a high, slow one. They add up to ${90^\circ}$.