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

The general solution for 5cos(3x)=-3 is x=(2.21429…)/3+(2pin)/3 ,x=-(2.21429…)/3+(2pin)/3

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5cos(3x)=-3

5 cos (3 x) = - 3

x = \frac{2.21429 \dots}{3} + \frac{2 πn}{3}, x = - \frac{2.21429 \dots}{3} + \frac{2 πn}{3}

Solution steps

5 cos (3 x) = - 3

Divide both sides by

5

cos (3 x) = - \frac{3}{5}

Apply trig inverse properties

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

Solve

3 x = arccos (- \frac{3}{5}) + 2 πn : x = \frac{arccos ( - \frac{3}{5} )}{3} + \frac{2 πn}{3}

Solve

3 x = - arccos (- \frac{3}{5}) + 2 πn : x = - \frac{arccos ( - \frac{3}{5} )}{3} + \frac{2 πn}{3}

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

Show solutions in decimal form

x = \frac{2.21429 \dots}{3} + \frac{2 πn}{3}, x = - \frac{2.21429 \dots}{3} + \frac{2 πn}{3}

2 sin^{3} (x) + sin^{2} (x) = 0

2 sin^{2} (θ) = sin (θ), 0 \leq θ \leq 2 π

cos (2 x) = cos (4 x)

14 sin^{2} (x) - 7 sin (x) = 0

cos (α) = \frac{2}{3}

What is the general solution for 5cos(3x)=-3 ?
The general solution for 5cos(3x)=-3 is x=(2.21429…)/3+(2pin)/3 ,x=-(2.21429…)/3+(2pin)/3