Question

Solve the equation on the

Answer 1

Answer:

see explanation

Step-by-step explanation:

note that (sinx)² = sin²x

4sin²x - 2 = 0 ( add 2 to both sides )

4sin²x = 2 ( divide both sides by 4 )

sin²x = $\frac{2}{4}$ = $\frac{1}{2}$ ( take square root of both sides )

sinx = ± $\sqrt{\frac{1}{2} }$ = ± $\frac{1}{\sqrt{2} }$

sinx = $\frac{1}{\sqrt{2} }$ ( sinx > 0 , then x is in 1st/2nd quadrants )

x = $\frac{\pi }{4}$ , π - $\frac{\pi }{4}$ = $\frac{3\pi }{4}$

sinx = - $\frac{1}{\sqrt{2} }$ ( sinx < 0 , then x is in 3rd/4th quadrants )

x = π + $\frac{\pi }{4}$ = $\frac{5\pi }{4}$ , 2π - $\frac{\pi }{4}$ = $\frac{7\pi }{4}$

solutions are

$\frac{\pi }{4}$ , $\frac{3\pi }{4}$ , $\frac{5\pi }{4}$ , $\frac{7\pi }{4}$ in the interval [ 0, 2π )