What is the general solution for 2sin^2(x)-15sin(x)+7=0 ?

The general solution for 2sin^2(x)-15sin(x)+7=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2sin^2(x)-15sin(x)+7=0

2 sin^{2} (x) - 15 sin (x) + 7 = 0

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

Solution steps

2 sin^{2} (x) - 15 sin (x) + 7 = 0

Solve by substitution

sin (x) = 7, sin (x) = \frac{1}{2}

sin (x) = 7 :

No Solution

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

Combine all the solutions

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

x, cos (x) = \frac{- 3}{5}

4 sin^{2} (x) = 4 cos (x) + 5

sin (3 x - \frac{π}{5}) = 0 - 1

2 + sin (3 θ) = 3 - sin (3 θ)

sin (2 x) = - cos (3 x) + sin (4 x)

What is the general solution for 2sin^2(x)-15sin(x)+7=0 ?
The general solution for 2sin^2(x)-15sin(x)+7=0 is x= pi/6+2pin,x=(5pi)/6+2pin