What is the general solution for 2sin^2(x)+9sin(x)-5=0 ?

The general solution for 2sin^2(x)+9sin(x)-5=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2sin^2(x)+9sin(x)-5=0

2 sin^{2} (x) + 9 sin (x) - 5 = 0

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

Solution steps

2 sin^{2} (x) + 9 sin (x) - 5 = 0

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = - 5

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

sin (x) = - 5 :

No Solution

Combine all the solutions

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

10 cos (2 x) = 0

6 cos^{2} (θ) sin (θ) - 3 cos^{2} (θ) = 0

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

2 + 7 cos (θ) = 4 sin^{2} (θ)

x, sinh (x) = 0

What is the general solution for 2sin^2(x)+9sin(x)-5=0 ?
The general solution for 2sin^2(x)+9sin(x)-5=0 is x= pi/6+2pin,x=(5pi)/6+2pin