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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 #4: I am an angle x such that 0\le x <2\pi . I satisfy the equation \sin 2x - \sin x=0 . What angle am I?


Lastly, we can use the half and double angle formulas to solve trigonometric equations.

Example A

Solve \tan 2x+\tan x=0 when 0\le x <2\pi .

Solution: Change \tan 2x and simplify.

\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

Set each factor equal to zero and solve.

&\ \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}

Example B

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

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

2\cos \frac{x}{2}+1 &=0 \\2\cos \frac{x}{2} &=-1 \\\cos \frac{x}{2} &= -\frac{1}{2}

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

\frac{x}{2} &=\frac{2\pi}{3},\frac{4\pi}{3} \\&=\frac{4\pi}{3}, \frac{8\pi}{3}

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

Example C

Solve 4\sin x \cos x = \sqrt{3} for 0\le x < 2\pi .

Solution: Pull a 2 out of the left-hand side and use the \sin 2x formula.

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}

Concept Problem Revisit

Use the double angle formula and simplify.

\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}

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

Guided Practice

Solve the following equations for 0\le x <2\pi .

1. \sin \frac{x}{2}=-1

2. \cos 2x-\cos x=0


1. \sin \frac{x}{2} &=-1 \\\frac{x}{2} &=\frac{3\pi}{2} \\x &=3\pi

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

2. \cos 2x - \cos x &=0 \\2\cos ^2x-\cos x -1 &=0 \\(2\cos x -1)(\cos x +1) &=0

Set each factor equal to zero and solve.

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 \\


Solve the following equations for 0\le x < 2\pi .

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

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