What is the general solution for 4cos(θ)+1=0 ?

The general solution for 4cos(θ)+1=0 is θ=1.82347…+2pin,θ=-1.82347…+2pin

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4cos(θ)+1=0

4 cos (θ) + 1 = 0

θ = 1.82347 \dots + 2 πn, θ = - 1.82347 \dots + 2 πn

Degrees

θ = 104.47751 \dots^{\circ} + 36 0^{\circ} n, θ = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (θ) + 1 = 0

Move

1

to the right side

4 cos (θ) + 1 = 0

Subtract

1

from both sides

4 cos (θ) + 1 - 1 = 0 - 1

Simplify

4 cos (θ) = - 1

4 cos (θ) = - 1

Divide both sides by

4

4 cos (θ) = - 1

Divide both sides by

4

\frac{4 cos ( θ )}{4} = \frac{- 1}{4}

Simplify

cos (θ) = - \frac{1}{4}

cos (θ) = - \frac{1}{4}

Apply trig inverse properties

cos (θ) = - \frac{1}{4}

General solutions for

cos (θ) = - \frac{1}{4}

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

θ = arccos (- \frac{1}{4}) + 2 πn, θ = - arccos (- \frac{1}{4}) + 2 πn

θ = arccos (- \frac{1}{4}) + 2 πn, θ = - arccos (- \frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = 1.82347 \dots + 2 πn, θ = - 1.82347 \dots + 2 πn

cos (a) = \frac{3}{- 6}

cos (x + 5) = \frac{1}{2} 3, 0 \leq x \leq 2 π

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

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

- 4 arctan (x) = π

What is the general solution for 4cos(θ)+1=0 ?
The general solution for 4cos(θ)+1=0 is θ=1.82347…+2pin,θ=-1.82347…+2pin