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

The general solution for 2sin^2(x)-1=0,-pi<= x<= pi is x= pi/4 ,x=(3pi)/4 ,x=-(3pi)/4 ,x=-pi/4

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

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

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

Solution steps

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

Solve by substitution

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

sin (x) = \frac{1}{2}, - π \leq x \leq π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, - π \leq x \leq π : x = - \frac{3 π}{4}, x = - \frac{π}{4}

Combine all the solutions

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

sin (θ) = - 0.22

\frac{3}{2} sec (θ) + 3 cos (θ) = 3

2 sec (x) cos^{2} (x) = csc^{2} (x) - cot^{2} (x)

1 = 7 cos (\frac{π}{3} t)

12 cos^{2} (x) + 2 cos (x) - 2 = 0

What is the general solution for 2sin^2(x)-1=0,-pi<= x<= pi ?
The general solution for 2sin^2(x)-1=0,-pi<= x<= pi is x= pi/4 ,x=(3pi)/4 ,x=-(3pi)/4 ,x=-pi/4