Solve equations of the form sin(x)=a - High school Trigonometry Exercises

Solve the following equations.

\sin\left(x\right) = \dfrac{-1}{2}

x = \left(\dfrac{4\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{7\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{\pi}{2}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{3\pi}{2}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{5\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{7\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{11\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z}

Looking at the unit circle, we can see that the function \sin\left(x\right) has a value of \dfrac{-1}{2} at \dfrac{7\pi}{6} and \dfrac{11\pi}{6}.

Since the sine function is periodic of period 2\pi, the full solution set is:

\left\{ x = \left(\dfrac{7\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z} \right\} and \left\{ x = \left(\dfrac{11\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z} \right\}

\sin\left(x\right) = \dfrac{-\sqrt{2}}{2}

x = \left(\dfrac{5\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{7\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z}

x = \dfrac{\pi}{2}\left(2n + 1\right), n \in \mathbb{Z}\\

x = \left(\dfrac{\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{2\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{3\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{5\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z}

Looking at the unit circle, we can see that the function \sin\left(x\right) has a value of \dfrac{-\sqrt{2}}{2} at \dfrac{5\pi}{4} and \dfrac{7\pi}{4}.

Since the sine function is periodic of period 2\pi, the full solution set is:

\left\{ x = \left(\dfrac{5\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z} \right\} and \left\{ x = \left(\dfrac{7\pi}{4}\right)+2n\pi, \ n \in \mathbb{Z} \right\}

\sin\left(x\right) = \dfrac{-\sqrt{3}}{2}

x = \left(\dfrac{2\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{5\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{4\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{5\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{7\pi}{6}\right)+2n\pi, n \in \mathbb{Z} or x = \left(\dfrac{11\pi}{6}\right)+2n\pi, n \in \mathbb{Z}

x = \left(\dfrac{\pi}{6}\right)+2n\pi, n \in \mathbb{Z} or x = \left(\dfrac{5\pi}{6}\right)+2n\pi, n \in \mathbb{Z}

Looking at the unit circle, we can see that the function \sin\left(x\right) has a value of \dfrac{-\sqrt{3}}{2} at \dfrac{4\pi}{3} and \dfrac{5\pi}{3}.

Since the sine function is periodic of period 2\pi, the full solution set is:

\left\{ x = \left(\dfrac{4\pi}{3}\right)+2n\pi, n \in \mathbb{Z} \right\} and \left\{x = \left(\dfrac{5\pi}{3}\right)+2n\pi, n \in \mathbb{Z} \right\}

\sin\left(2x\right) = -1

x = \left(\dfrac{3\pi}{2}\right)+2n\pi, \ n \in \mathbb{Z}

x = \pi\left(n + \dfrac{1}{2}\right), n \in \mathbb{Z}

x = \left(\dfrac{3\pi}{4}\right)+n\pi, \ n \in \mathbb{Z}

x = \dfrac{\pi}{2}\left(n + \dfrac{3}{2}\right), n \in \mathbb{Z}

Looking at the unit circle, we can see that the function \sin\left(x\right) has a value of -1 at \dfrac{3\pi}{2}

Since the sine function is periodic of period 2\pi, the full solution set is:

2x = \dfrac{3\pi}{2} + 2\pi n, \ n \in \mathbb{Z}

x = \dfrac{3\pi}{4} + \pi n,\ n \in \mathbb{Z}

\left\{ x = \left(\dfrac{3\pi}{4}\right)+n\pi, \ n \in \mathbb{Z} \right\}

\sin\left(2x\right) = \dfrac{-1}{2}

x = \left(\dfrac{7\pi}{12}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{11\pi}{12}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{7\pi}{6}\right) + 2\pi n, n \in \mathbb{Z}\\ or x = \left(\dfrac{11\pi}{6}\right) + 2\pi n, n \in \mathbb{Z}

x = \left(\dfrac{7\pi}{12}\right) + \pi n, n \in \mathbb{Z} or x = \left(\dfrac{11\pi}{12}\right) + \pi n, n \in \mathbb{Z}

x = \left(\dfrac{5\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{7\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z}

Looking at the unit circle, we can see that the function \sin\left(x\right) has a value of \dfrac{-1}{2} at \dfrac{7\pi}{6} and \dfrac{11\pi}{6}.

Since the sine function is periodic of period 2\pi, the full solution set is:

2x = \dfrac{7\pi}{6} + 2\pi n, n \in \mathbb{Z}\\x = \dfrac{7\pi}{12} + \pi n, n \in \mathbb{Z}

2x = \dfrac{11\pi}{6} + 2\pi n, n \in \mathbb{Z}\\x = \dfrac{11\pi}{12} + \pi n, n \in \mathbb{Z}

\left\{ x = \left(\dfrac{7\pi}{12}\right)+n\pi, \ n \in \mathbb{Z} \right\} and \left\{ x = \left(\dfrac{11\pi}{12}\right)+n\pi, \ n \in \mathbb{Z} \right\}

\sin\left(x\right) = 0

x = \dfrac{\pi}{2}\left(4n + 1\right), \ n \in \mathbb{Z}

x = \dfrac{\pi}{2}\left(2n+1\right), \ n \in \mathbb{Z}

x = \pi n, \ n \in \mathbb{Z}

x = 2\pi n, \ n \in \mathbb{Z}

Looking at the unit circle, we can see that the function \sin\left(x\right) has a value of 0 at 0 and \pi.

Since the sine function is periodic of period 2\pi, the full solution set is:

x = \pi n, \ n \in \mathbb{Z}

\sin\left(x\right) = \dfrac{\sqrt{3}}{2}

x = \left(\dfrac{4\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{7\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{5\pi}{6}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{2\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{4\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z}

x = \left(\dfrac{\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} or x = \left(\dfrac{2\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z}

Looking at the unit circle, we can see that the function \sin\left(x\right) has a value of \dfrac{\sqrt{3}}{2} at \dfrac{\pi}{3} and \dfrac{2\pi}{3}.

Since the sine function is periodic of period 2\pi, the full solution set is:

\left\{ x = \left(\dfrac{\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} \right\} and \left\{ x = \left(\dfrac{2\pi}{3}\right)+2n\pi, \ n \in \mathbb{Z} \right\}