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

The general solution for 2sin^2(x)=1,0<= x<= 2pi is x= pi/4 ,x=(3pi)/4 ,x=(5pi)/4 ,x=(7pi)/4

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

2 sin^{2} (x) = 1, 0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Solution steps

2 sin^{2} (x) = 1, 0 \leq x \leq 2 π

Solve by substitution

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

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

0 = \frac{1}{5} sin (5 t)

tan^{2} (x) - 1 = 0, 0 \leq x < 2 π

6 sin^{2} (x) - 5 sin (x) + 1 = 0

cos (θ) + 1 = sin (θ)

2 sec^{2} (x) - 2 = 0

What is the general solution for 2sin^2(x)=1,0<= x<= 2pi ?
The general solution for 2sin^2(x)=1,0<= x<= 2pi is x= pi/4 ,x=(3pi)/4 ,x=(5pi)/4 ,x=(7pi)/4