What is the general solution for 5cos(θ)-3=0 ?

The general solution for 5cos(θ)-3=0 is θ=0.92729…+2pin,θ=2pi-0.92729…+2pin

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5cos(θ)-3=0

5 cos (θ) - 3 = 0

θ = 0.92729 \dots + 2 πn, θ = 2 π - 0.92729 \dots + 2 πn

Degrees

θ = 53.13010 \dots^{\circ} + 36 0^{\circ} n, θ = 306.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 cos (θ) - 3 = 0

Move

3

to the right side

5 cos (θ) - 3 = 0

Add

3

to both sides

5 cos (θ) - 3 + 3 = 0 + 3

Simplify

5 cos (θ) = 3

5 cos (θ) = 3

Divide both sides by

5

5 cos (θ) = 3

Divide both sides by

5

\frac{5 cos ( θ )}{5} = \frac{3}{5}

Simplify

cos (θ) = \frac{3}{5}

cos (θ) = \frac{3}{5}

Apply trig inverse properties

cos (θ) = \frac{3}{5}

General solutions for

cos (θ) = \frac{3}{5}

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

θ = arccos (\frac{3}{5}) + 2 πn, θ = 2 π - arccos (\frac{3}{5}) + 2 πn

θ = arccos (\frac{3}{5}) + 2 πn, θ = 2 π - arccos (\frac{3}{5}) + 2 πn

Show solutions in decimal form

θ = 0.92729 \dots + 2 πn, θ = 2 π - 0.92729 \dots + 2 πn

(cos (x)) (sin (x)) = - 1

- 2 tan (x) sec^{2} (x) = 0

2 - sin (x) = sin (x)

cos^{2} (t) = \frac{1}{2}

2 cos^{2} (θ) - 1 = sin (θ), 0^{\circ} \leq θ \leq 18 0^{\circ}

What is the general solution for 5cos(θ)-3=0 ?
The general solution for 5cos(θ)-3=0 is θ=0.92729…+2pin,θ=2pi-0.92729…+2pin