Question

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

Answer 1

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=(1)/(2) \end{gathered}`

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

`\begin{gathered} \text{sin(x)}=(1)/(2) \\ x=\text{sin}^(-1)((1)/(2)) \\ x=30^0=(\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=(5\pi)/(6) \end{gathered}`

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