What is the general solution for -6cos^2(θ)=0 ?

The general solution for -6cos^2(θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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-6cos^2(θ)=0

- 6 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

- 6 cos^{2} (θ) = 0

Divide both sides by

- 6

- 6 cos^{2} (θ) = 0

Divide both sides by

- 6

- 6 cos^{2} (θ) = 0

Divide both sides by

- 6

\frac{- 6 cos ^{2} ( θ )}{- 6} = \frac{0}{- 6}

Simplify

cos^{2} (θ) = 0

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

arctan (x) = 60

tan (x) = - \frac{24}{7} tan (2 x)

tan (x) = \frac{13}{9}

cos (A) = \frac{6}{7 \cdot 21}

2 = 4 cos (3 x + 1)

What is the general solution for -6cos^2(θ)=0 ?
The general solution for -6cos^2(θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin