What is the general solution for -4cos(3θ)=1 ?

The general solution for -4cos(3θ)=1 is θ=(1.82347…)/3+(2pin)/3 ,θ=-(1.82347…)/3+(2pin)/3

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-4cos(3θ)=1

- 4 cos (3 θ) = 1

θ = \frac{1.82347 \dots}{3} + \frac{2 πn}{3}, θ = - \frac{1.82347 \dots}{3} + \frac{2 πn}{3}

Solution steps

- 4 cos (3 θ) = 1

Divide both sides by

- 4

cos (3 θ) = - \frac{1}{4}

Apply trig inverse properties

3 θ = arccos (- \frac{1}{4}) + 2 πn, 3 θ = - arccos (- \frac{1}{4}) + 2 πn

Solve

3 θ = arccos (- \frac{1}{4}) + 2 πn : θ = \frac{arccos ( - \frac{1}{4} )}{3} + \frac{2 πn}{3}

Solve

3 θ = - arccos (- \frac{1}{4}) + 2 πn : θ = - \frac{arccos ( - \frac{1}{4} )}{3} + \frac{2 πn}{3}

θ = \frac{arccos ( - \frac{1}{4} )}{3} + \frac{2 πn}{3}, θ = - \frac{arccos ( - \frac{1}{4} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

θ = \frac{1.82347 \dots}{3} + \frac{2 πn}{3}, θ = - \frac{1.82347 \dots}{3} + \frac{2 πn}{3}

960 = 480 (1 - cos (7.5 x))

8 = 8 tan (θ)

2 sin (2 x) - 3 cos (x) = 0

4 cos (θ) = 2 cos (θ)

2 cos (θ) = - 2

What is the general solution for -4cos(3θ)=1 ?
The general solution for -4cos(3θ)=1 is θ=(1.82347…)/3+(2pin)/3 ,θ=-(1.82347…)/3+(2pin)/3