Question

All the solutions of the equation $\sin \theta =\frac{1}{2}$are ____.

Short answer

All the solutions of the equation $\sin \theta =\frac{1}{2}$are$\left\{\theta |\theta =\frac{\mathrm{\pi }}{6}+2\mathrm{\pi k},\mathrm{\theta }=\frac{5\mathrm{\pi }}{6}+2\mathrm{\pi k},\mathrm{k}\mathrm{is}\mathrm{any}\mathrm{integer}\right\}$.

Step-by-step solution

Step 1. Given Information

All the solutions of the equation $\sin \theta =\frac{1}{2}$are ____.

Step 2. The period of the sine function is 2π.

In the interval $[0,2\mathrm{\pi })$, there are two angles $\theta$for which

So all the solution of $\sin \theta =\frac{1}{2}$are width="341">θ|θ=π6+2πk,θ=5π6+2πk,kisanyinteger