What is the general solution for 5cos(3x)=4 ?

The general solution for 5cos(3x)=4 is x=(0.64350…)/3+(2pin)/3 ,x=(2pi)/3-(0.64350…)/3+(2pin)/3

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5cos(3x)=4

5 cos (3 x) = 4

x = \frac{0.64350 \dots}{3} + \frac{2 πn}{3}, x = \frac{2 π}{3} - \frac{0.64350 \dots}{3} + \frac{2 πn}{3}

Solution steps

5 cos (3 x) = 4

Divide both sides by

5

cos (3 x) = \frac{4}{5}

Apply trig inverse properties

3 x = arccos (\frac{4}{5}) + 2 πn, 3 x = 2 π - arccos (\frac{4}{5}) + 2 πn

Solve

3 x = arccos (\frac{4}{5}) + 2 πn : x = \frac{arccos ( \frac{4}{5} )}{3} + \frac{2 πn}{3}

Solve

3 x = 2 π - arccos (\frac{4}{5}) + 2 πn : x = \frac{2 π}{3} - \frac{arccos ( \frac{4}{5} )}{3} + \frac{2 πn}{3}

x = \frac{arccos ( \frac{4}{5} )}{3} + \frac{2 πn}{3}, x = \frac{2 π}{3} - \frac{arccos ( \frac{4}{5} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

x = \frac{0.64350 \dots}{3} + \frac{2 πn}{3}, x = \frac{2 π}{3} - \frac{0.64350 \dots}{3} + \frac{2 πn}{3}

tan (\frac{θ}{2} + \frac{π}{4}) = - 1

2 cos^{2} (θ) - 11 cos (θ) + 5 = 0

4 sin (2 x) - 1 = 0

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

x, cos (x) + sin (x) = 0

What is the general solution for 5cos(3x)=4 ?
The general solution for 5cos(3x)=4 is x=(0.64350…)/3+(2pin)/3 ,x=(2pi)/3-(0.64350…)/3+(2pin)/3