Question

solve ( 2 sin ^ { 2 } ( x ) + 3 sin ( x ) + 1 = 0 ) for all solutions ( 0 leq x < 2 pi ).

( x = )

give your answers as exact values in a list separated by commas.

question help: video

Explanation:

Step1: Let \( t = \sin(x) \)

The equation \( 2\sin^{2}(x)+3\sin(x)+1 = 0 \) becomes \( 2t^{2}+3t + 1=0 \)

Step2: Factor the quadratic equation

We factor \( 2t^{2}+3t + 1\) as \((2t + 1)(t+1)=0\)

Step3: Solve for \( t \)

Set each factor equal to zero:

• For \( 2t+1 = 0\), we get \(t=-\frac{1}{2}\)
• For \(t + 1=0\), we get \(t=-1\)

Step4: Substitute back \( t=\sin(x) \)

• When \(\sin(x)=-\frac{1}{2}\), \(x=\frac{7\pi}{6}\) or \(x=\frac{11\pi}{6}\) (since \(\sin(x)\) is negative in the third and fourth quadrants and \(\sin(\frac{\pi}{6})=\frac{1}{2}\))
• When \(\sin(x)=-1\), \(x=\frac{3\pi}{2}\)

Answer:

\(\frac{3\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}\)