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Popular Trigonometry >

cos^2(θ)= 1/3

Solution

cos^{2} (θ) = \frac{1}{3}

Solution

θ = 0.95531 \dots + 2 πn, θ = 2 π - 0.95531 \dots + 2 πn, θ = 2.18627 \dots + 2 πn, θ = - 2.18627 \dots + 2 πn

+1

Degrees

θ = 54.73561 \dots^{\circ} + 36 0^{\circ} n, θ = 305.26438 \dots^{\circ} + 36 0^{\circ} n, θ = 125.26438 \dots^{\circ} + 36 0^{\circ} n, θ = - 125.26438 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) = \frac{1}{3}

Solve by substitution

cos^{2} (θ) = \frac{1}{3}

Let:

cos (θ) = u

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3} : u = \frac{1}{3}, u = - \frac{1}{3}

u^{2} = \frac{1}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{3}, cos (θ) = - \frac{1}{3}

cos (θ) = \frac{1}{3}, cos (θ) = - \frac{1}{3}

cos (θ) = \frac{1}{3} : θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (θ) = \frac{1}{3}

Apply trig inverse properties

cos (θ) = \frac{1}{3}

General solutions for

cos (θ) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (θ) = - \frac{1}{3} : θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

cos (θ) = - \frac{1}{3}

Apply trig inverse properties

cos (θ) = - \frac{1}{3}

General solutions for

cos (θ) = - \frac{1}{3}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

Combine all the solutions

θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn, θ = arccos (- \frac{1}{3}) + 2 πn, θ = - arccos (- \frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = 0.95531 \dots + 2 πn, θ = 2 π - 0.95531 \dots + 2 πn, θ = 2.18627 \dots + 2 πn, θ = - 2.18627 \dots + 2 πn

Graph

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