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

The general solution for -2sin^2(x)-cos(x)=-1 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin

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

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

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solution steps

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

Subtract

- 1

from both sides

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

cos (x) = 1 : x = 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

tan^{2} (x) = 1 + sec (x)

8 sin^{2} (x) = 7

- 6 arcsin (3 x) = π

tan^{2} (p) = tan (p)

sin (x) + cos (x) = \frac{2}{3}

What is the general solution for -2sin^2(x)-cos(x)=-1 ?
The general solution for -2sin^2(x)-cos(x)=-1 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin