What is the general solution for 2sin^2(x)=4sin(x)+6,(0,2pi) ?

The general solution for 2sin^2(x)=4sin(x)+6,(0,2pi) is x=(3pi)/2

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2sin^2(x)=4sin(x)+6,(0,2pi)

2 sin^{2} (x) = 4 sin (x) + 6, (0, 2 π)

x = \frac{3 π}{2}

Solution steps

2 sin^{2} (x) = 4 sin (x) + 6, (0, 2 π)

Solve by substitution

sin (x) = 3, sin (x) = - 1

sin (x) = 3, (0, 2 π) :

No Solution

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

Combine all the solutions

x = \frac{3 π}{2}

2 cos (3 x + 2 0^{\circ}) = 2

cot^{2} (3 x) - 2 cot (3 x) + 1 = 0

- \frac{tanh ( n x )}{cosh ( n x )} = 0

sin (φ) = 0

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

What is the general solution for 2sin^2(x)=4sin(x)+6,(0,2pi) ?
The general solution for 2sin^2(x)=4sin(x)+6,(0,2pi) is x=(3pi)/2