What is the general solution for cos^2(θ)=0 ?

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

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cos^2(θ)=0

cos^{2} (θ) = 0

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

cos (7 x) cos (2 x) + sin (7 x) sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}, 0 \leq x \leq 2 π

8 sin^{2} (θ) - 6 = 0

3 sin (x) - cos (x) = 0

3 cos (x) tan^{2} (x) = cos (x)

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