What is the general solution for 2sin^2(x)+7sin(x)-4=0 ?

The general solution for 2sin^2(x)+7sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2sin^2(x)+7sin(x)-4=0

2 sin^{2} (x) + 7 sin (x) - 4 = 0

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

Solution steps

2 sin^{2} (x) + 7 sin (x) - 4 = 0

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = - 4

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

sin (x) = - 4 :

No Solution

Combine all the solutions

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

cos (\frac{5 π}{4} + x) + sin (\frac{5 π}{4} - x) = 0

cos (2 x) + sin (2 x) = 1

θ, sin (θ) = \frac{1}{2}

sin (x + \frac{3 π}{2}) - cos^{2} (x) = 0

6 cos (x) - 6 sin (x) = 0

What is the general solution for 2sin^2(x)+7sin(x)-4=0 ?
The general solution for 2sin^2(x)+7sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin