What is the general solution for 2sin^2(x)-17sin(x)+8=0 ?

The general solution for 2sin^2(x)-17sin(x)+8=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2sin^2(x)-17sin(x)+8=0

2 sin^{2} (x) - 17 sin (x) + 8 = 0

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

Solution steps

2 sin^{2} (x) - 17 sin (x) + 8 = 0

Solve by substitution

sin (x) = 8, sin (x) = \frac{1}{2}

sin (x) = 8 :

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

4 cos^{2} (t) sin (t) - 2 cos (t) sin (t) = 0

3 (1 - sin (θ)) = 2 cos^{2} (θ)

sin (x) = \frac{5}{5}

tan (θ) = 0.11246

cos (2 θ) - sin^{2} (θ) = 1

What is the general solution for 2sin^2(x)-17sin(x)+8=0 ?
The general solution for 2sin^2(x)-17sin(x)+8=0 is x= pi/6+2pin,x=(5pi)/6+2pin