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

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

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

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

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

Solution steps

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

Solve by substitution

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

sin (x) = 2 :

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

sin (x + \frac{π}{3}) + sin (x - \frac{π}{3}) = 1

sec^{2} (x) = 4

1 + 2 cos (x) = 0

sin (2 x) = 3 cos (x)

arctan (x) = 1

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