Question

Solve the equation for exact solutions over the interval [0, 2 π).2sin^2x=sin x

Answer 1

Answer:

π/6 and 5π/6 radians

Explanations:

Given the trigonometry expression

$2\sin ^2x=\sin x$

Let P = sin(x) to have:

$\begin{gathered} 2P^2=P \\ 2P=1 \\ P=\frac{1}{2} \end{gathered}$

Substitute P = sin(x) into the resulting expression;

$\begin{gathered} \text{sin(x)}=\frac{1}{2} \\ x=\text{sin}^{-1}(\frac{1}{2}) \\ x=30^0=\frac{\pi}{6} \end{gathered}$

Since the equation is over the interval [0, 2 π) and sin(theta) is positive in the 2nd quadrant, hence other angles will be expressed as:

$\begin{gathered} x_2=180-\theta \\ x_2=180-30 \\ x_2=150^0=\frac{5\pi}{6} \end{gathered}$

Therefore the exact solutions over the interval [0, 2 π) are π/6 and 5π/6 radians respectively