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

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

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

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

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

Solution steps

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

Rewrite using trig identities

- 3 + 3 sin^{2} (x) = 0

Solve by substitution

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

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

Combine all the solutions

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

cos (3 x) \cdot 3 = 0

16 sec^{2} (θ) - 25 = 0

2 csc^{2} (x) - 4 = 0

2 sin^{2} (θ) - 3 sin (θ) - 2 = 0, 0 \leq θ \leq 2 π

3 sin (2 θ) = 2 cos (2 θ), 0^{\circ} \leq 2 θ < 72 0^{\circ}

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