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

The general solution for 2sin^2(x)-3sin(x)=-1,0<= x<= 2pi is x= pi/2 ,x= pi/6 ,x=(5pi)/6

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2sin^2(x)-3sin(x)=-1,0<= x<= 2pi

2 sin^{2} (x) - 3 sin (x) = - 1, 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

2 sin^{2} (x) - 3 sin (x) = - 1, 0 \leq x \leq 2 π

Solve by substitution

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

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

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

x = \frac{π}{2}, x = \frac{π}{6}, x = \frac{5 π}{6}

5 cos (x) + 23 = cos (x)

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

tan^{2} (θ) = 3 tan (θ)

4 sin^{2} (2) cot (x) - 2 = 0

5 sin (2 θ) + 2 sin (θ) = 0

What is the general solution for 2sin^2(x)-3sin(x)=-1,0<= x<= 2pi ?
The general solution for 2sin^2(x)-3sin(x)=-1,0<= x<= 2pi is x= pi/2 ,x= pi/6 ,x=(5pi)/6