What is the general solution for 4cos(θ)+3=0 ?

The general solution for 4cos(θ)+3=0 is θ=2.41885…+2pin,θ=-2.41885…+2pin

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4cos(θ)+3=0

4 cos (θ) + 3 = 0

θ = 2.41885 \dots + 2 πn, θ = - 2.41885 \dots + 2 πn

Degrees

θ = 138.59037 \dots^{\circ} + 36 0^{\circ} n, θ = - 138.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (θ) + 3 = 0

Move

3

to the right side

4 cos (θ) + 3 = 0

Subtract

3

from both sides

4 cos (θ) + 3 - 3 = 0 - 3

Simplify

4 cos (θ) = - 3

4 cos (θ) = - 3

Divide both sides by

4

4 cos (θ) = - 3

Divide both sides by

4

\frac{4 cos ( θ )}{4} = \frac{- 3}{4}

Simplify

cos (θ) = - \frac{3}{4}

cos (θ) = - \frac{3}{4}

Apply trig inverse properties

cos (θ) = - \frac{3}{4}

General solutions for

cos (θ) = - \frac{3}{4}

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

θ = arccos (- \frac{3}{4}) + 2 πn, θ = - arccos (- \frac{3}{4}) + 2 πn

θ = arccos (- \frac{3}{4}) + 2 πn, θ = - arccos (- \frac{3}{4}) + 2 πn

Show solutions in decimal form

θ = 2.41885 \dots + 2 πn, θ = - 2.41885 \dots + 2 πn

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

cos (x) = \frac{5}{9}

sin^{2} (x) = - \frac{1}{2}

4 tan^{2} (θ) = 11 tan (θ) + 3

6 tan^{2} (x) - 2 cos^{2} (x) = cos (2 x)

What is the general solution for 4cos(θ)+3=0 ?
The general solution for 4cos(θ)+3=0 is θ=2.41885…+2pin,θ=-2.41885…+2pin