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

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

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

2 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

- 6 sin (t) + 6 cos (2 t) = 0

cos^{2} (x) = 4

- 2 tan (θ) - sec (θ) = 0

4 cos (x) = - sin^{2} (x) + 1, 0 \leq x \leq 2 π

2 \cdot sin (x) = tan (x)

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