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

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

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

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

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

Solution steps

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

Rewrite using trig identities

3 cos^{2} (x) = 0

Divide both sides by

3

cos (x) = 0

General solutions for

cos (x) = 0

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

1 = sinh (x)

15 cos^{2} (θ) + cos (θ) - 2 = 0

sin (θ) = \frac{9}{11}

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

3 sin^{2} (t) - 2 sin (t) - 1 = 0

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