What is the general solution for 4sin^2(x)-1=0,0<= x<= pi/2 ?

The general solution for 4sin^2(x)-1=0,0<= x<= pi/2 is x= pi/6

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4sin^2(x)-1=0,0<= x<= pi/2

4 sin^{2} (x) - 1 = 0, 0 \leq x \leq \frac{π}{2}

x = \frac{π}{6}

Solution steps

4 sin^{2} (x) - 1 = 0, 0 \leq x \leq \frac{π}{2}

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, 0 \leq x \leq \frac{π}{2} : x = \frac{π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x \leq \frac{π}{2} :

No Solution

Combine all the solutions

x = \frac{π}{6}

\frac{2}{3} = cos (x)

2 cos (\frac{x}{2}) - 3 = 0

cos (2 x) + 3 cos (x) = 1

cos (2 θ) - cos (θ) = 2

sin (θ) - cos^{2} (θ) sin (θ) = - 1

What is the general solution for 4sin^2(x)-1=0,0<= x<= pi/2 ?
The general solution for 4sin^2(x)-1=0,0<= x<= pi/2 is x= pi/6