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Solving Equations with Double-Angle Identities

Solve sine, cosine, and tangent of angles multiplied or divided by 2.

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Solving Trig Equations using Double and Half Angle Formulas

Trig Riddle: I am an angle x such that \begin{align*}0\le x <2\pi\end{align*}. I satisfy the equation \begin{align*}\sin 2x - \sin x=0\end{align*}. What angle am I?

Solve Trigonometric Equations

We can use the half and double angle formulas to solve trigonometric equations.

Let's solve the following trigonometric equations.

  1. Solve \begin{align*}\tan 2x+\tan x=0\end{align*} when \begin{align*}0\le x <2\pi\end{align*}.

Change \begin{align*}\tan 2x\end{align*} and simplify.

\begin{align*}\tan 2x + \tan x &=0 \\ \frac{2\tan x}{1-\tan ^2 x}+\tan x &=0 \\ 2\tan x +\tan x(1-\tan ^2x) &=0 \quad \rightarrow \text{Multiply everything by} \ 1-\tan^2x \text{ to eliminate denominator.}\\ 2\tan x +\tan x -\tan ^3 x &=0 \\ 3\tan x - \tan ^3 x &=0 \\ \tan x(3-\tan^2 x) &=0\end{align*}

Set each factor equal to zero and solve.

\begin{align*}&\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3-\tan ^2x =0 \\ &\ \ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ -\tan ^2 x =-3 \\ &\tan x=0 \qquad \qquad \qquad and \qquad \qquad \ \tan^2 x=3 \\ & \qquad x=0 \ and \ \pi \ \qquad \qquad \qquad \quad \qquad \tan x =\pm \sqrt{3} \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x =\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\end{align*}

  1. Solve \begin{align*}2\cos \frac{x}{2}+1=0\end{align*} when \begin{align*}0\le x<2\pi\end{align*}.

In this case, you do not have to use the half-angle formula. Solve for \begin{align*}\frac{x}{2}\end{align*}.

\begin{align*}2\cos \frac{x}{2}+1 &=0 \\ 2\cos \frac{x}{2} &=-1 \\ \cos \frac{x}{2} &= -\frac{1}{2}\end{align*}

Now, let’s find \begin{align*}\cos a = -\frac{1}{2}\end{align*} and then solve for \begin{align*}x\end{align*} by dividing by 2.

\begin{align*}\frac{x}{2} &=\frac{2\pi}{3},\frac{4\pi}{3} \\ &=\frac{4\pi}{3}, \frac{8\pi}{3}\end{align*}

Now, the second solution is not in our range, so the only solution is \begin{align*}x=\frac{4\pi}{3}\end{align*}.

  1. Solve \begin{align*}4\sin x \cos x = \sqrt{3}\end{align*} for \begin{align*}0\le x < 2\pi\end{align*}.

Pull a 2 out of the left-hand side and use the \begin{align*}\sin 2x\end{align*} formula.

\begin{align*}4\sin x \cos x &=\sqrt{3} \\ 2\cdot 2\sin x \cos x &=\sqrt{3} \\ 2 \cdot \sin 2x &=\sqrt{3} \\ \sin 2x &=\frac{\sqrt{3}}{2} \\ 2x &= \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3} \\ x &= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\end{align*}


Example 1

Earlier, you were asked to find the angle x, where \begin{align*}0\le x <2\pi\end{align*}, such that x satisfies the equation \begin{align*}\sin 2x - \sin x=0\end{align*}.

Use the double angle formula and simplify.

\begin{align*}\sin 2x - \sin x = 0\\ 2\sin x \cos x - \sin x = 0\\ \sin x(2\cos x - 1)= 0\\ \sin x = 0 \text OR \cos x = \frac{1}{2}\end{align*}

Under the constraint \begin{align*}0\le x <2\pi\end{align*}, \begin{align*}\sin x = 0\end{align*} when \begin{align*}x=0\end{align*} or when \begin{align*}x=\pi\end{align*}. Under this same constraint, \begin{align*}\cos x = \frac{1}{2}\end{align*} when \begin{align*}x=\frac{\pi}{3}\end{align*} or when \begin{align*}x=\frac{5\pi}{3}\end{align*}.

Example 2

Solve the following equation for \begin{align*}0\le x <2\pi\end{align*}.

\begin{align*}\sin \frac{x}{2}=-1\end{align*}

\begin{align*}\sin \frac{x}{2} &=-1 \\ \frac{x}{2} &=\frac{3\pi}{2} \\ x &=3\pi\end{align*}

From this we can see that there are no solutions within our interval.

Example 3

Solve the following equation for \begin{align*}0\le x <2\pi\end{align*}.

\begin{align*}\cos 2x-\cos x=0\end{align*}

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

Set each factor equal to zero and solve.

\begin{align*}2\cos x-1 &=0 \\ 2\cos x &=1 \qquad \qquad \qquad \qquad\cos x +1=0\\ \cos x &=\frac{1}{2} \qquad \qquad and \qquad \qquad \cos x =-1\\ x &=\frac{\pi}{3}, \frac{5\pi}{3} \qquad \qquad \qquad \qquad \quad \ x=\pi \\ \end{align*}


Solve the following equations for \begin{align*}0\le x < 2\pi\end{align*}.

  1. \begin{align*}\cos x -\cos \frac{1}{2}x=0\end{align*}
  2. \begin{align*}\sin 2x \cos x=\sin x\end{align*}
  3. \begin{align*}\cos 3x - \cos ^3x=3\sin ^2x\cos x\end{align*}
  4. \begin{align*}\tan 2x - \tan x =0\end{align*}
  5. \begin{align*}\cos 2x -\cos x =0\end{align*}
  6. \begin{align*}2\cos ^2\frac{x}{2}=1\end{align*}
  7. \begin{align*}\tan \frac{x}{2}=4\end{align*}
  8. \begin{align*}\cos \frac{x}{2}=1+\cos x\end{align*}
  9. \begin{align*}\sin 2x +\sin x=0\end{align*}
  10. \begin{align*}\cos ^2x-\cos 2x =0\end{align*}
  11. \begin{align*}\frac{\cos 2x}{\cos ^2x}=1\end{align*}
  12. \begin{align*}\cos 2x-1=\sin^2x\end{align*}
  13. \begin{align*}\cos 2x =\cos x\end{align*}
  14. \begin{align*}\sin 2x-\cos 2x =1\end{align*}
  15. \begin{align*}\sin^2x-2=\cos 2x\end{align*}
  16. \begin{align*}\cot x+\tan x=2\csc 2x\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.17. 

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