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=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin

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

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

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

cos (x) = 1 : x = 2 πn

Combine all the solutions

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

sin (4 x) = - 2 sin (2 x)

tan (x) = - 2

csc (x) = 2

4 cos^{2} (θ) = 1

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

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=(2pi)/3+2pin,x=(4pi)/3+2pin,x=2pin