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

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

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

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

x = \frac{3 π}{2}

Solution steps

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

Solve by substitution

sin (x) = \frac{3}{2}, sin (x) = - 1

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

No Solution

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

Combine all the solutions

x = \frac{3 π}{2}

- 15 sin (3 x) = 0

sin (θ) = - \frac{5}{3} cos (2 θ)

y, e^{c o s (x)} (1 - cos (y)) = c

cot (2 θ) = 1

cos (2 x) = 1 - sin^{2} (x)

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