Question

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.) 8 sin(2θ) − 2 sin(θ) = 0

Answer 1

Answer:

$\theta=k\pi, \hspace{3}k\in Z\\\\ or\\\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z$

Step-by-step explanation:

Factor constant terms:

$-2(-4sin(2\theta)+sin(\theta))=0$

Divide both sides by -2:

$sin(\theta)-4sin(2 \theta)=0$

Expand trigonometric functions using the fact:

$sin(2 \theta) =2 sin(\theta) cos(\theta)$

So:

$sin(\theta) -8sin(\theta)cos(\theta)=0$

Factor sin(x) and constant terms and multiply both sides by -1:

$sin(\theta) (8cos(\theta)-1)=0$

Split into two equations:

$(1)=8cos(\theta)-1=0\\\\(2)=sin(\theta)=0$

For (1)

Add 1 to both sides and divide both sides by 8:

$cos(\theta)=\frac{1}{8}$

Take the inverse cosine of both sides:

$\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z$

For (2)

Simply take the inverse sine of both sides

$\theta = k \pi, \hspace{3}k\in Z$

Therefore, the solutions are given by:

$\theta=k\pi, \hspace{3}k\in Z\\\\ or\\ \theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z$