What is the general solution for 1+4sin^2(x)-5sin(x)+cos(2x)=0 ?

The general solution for 1+4sin^2(x)-5sin(x)+cos(2x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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1+4sin^2(x)-5sin(x)+cos(2x)=0

1 + 4 sin^{2} (x) - 5 sin (x) + cos (2 x) = 0

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

Solution steps

1 + 4 sin^{2} (x) - 5 sin (x) + cos (2 x) = 0

Rewrite using trig identities

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

cos (x) = \frac{9}{10}

cos^{2} (x) + 4 cos (x) + 4 = 0

2 cos (x) = tan (x)

sin (x) + sin^{2} (x) = 1

sin^{3} (a) = 3 sin (a) - 4 sin^{3} (a)

What is the general solution for 1+4sin^2(x)-5sin(x)+cos(2x)=0 ?
The general solution for 1+4sin^2(x)-5sin(x)+cos(2x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin