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

The general solution for -2sin^2(x)+cos(x)+1=0 is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin

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

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

sin (x) tan (x) = 2 tan (x)

tan^{2} (x) - 2 tan (x) = 0

sin^{2} (θ) = 1

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

4 sin (x) cos (x) + cos (x) = 0

What is the general solution for -2sin^2(x)+cos(x)+1=0 ?
The general solution for -2sin^2(x)+cos(x)+1=0 is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin