What is the general solution for cos(3x)= 1/5 ?

The general solution for cos(3x)= 1/5 is x=(1.36943…)/3+(2pin)/3 ,x=(2pi)/3-(1.36943…)/3+(2pin)/3

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cos(3x)= 1/5

cos (3 x) = \frac{1}{5}

x = \frac{1.36943 \dots}{3} + \frac{2 πn}{3}, x = \frac{2 π}{3} - \frac{1.36943 \dots}{3} + \frac{2 πn}{3}

Solution steps

cos (3 x) = \frac{1}{5}

Apply trig inverse properties

3 x = arccos (\frac{1}{5}) + 2 πn, 3 x = 2 π - arccos (\frac{1}{5}) + 2 πn

Solve

3 x = arccos (\frac{1}{5}) + 2 πn : x = \frac{arccos ( \frac{1}{5} )}{3} + \frac{2 πn}{3}

Solve

3 x = 2 π - arccos (\frac{1}{5}) + 2 πn : x = \frac{2 π}{3} - \frac{arccos ( \frac{1}{5} )}{3} + \frac{2 πn}{3}

x = \frac{arccos ( \frac{1}{5} )}{3} + \frac{2 πn}{3}, x = \frac{2 π}{3} - \frac{arccos ( \frac{1}{5} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

x = \frac{1.36943 \dots}{3} + \frac{2 πn}{3}, x = \frac{2 π}{3} - \frac{1.36943 \dots}{3} + \frac{2 πn}{3}

cos^{2} (x) - 5 cos (x) + 6 = 0

cot^{2} (\frac{x}{2}) = 3, 0, 0 \leq x \leq 360

2 sin^{2} (x) + 5 sin (x) - 3 = 0, 0 \leq x \leq 2 π

sin (x) = \frac{123 sin ( 7 0 ^{\circ} )}{184.5}

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

What is the general solution for cos(3x)= 1/5 ?
The general solution for cos(3x)= 1/5 is x=(1.36943…)/3+(2pin)/3 ,x=(2pi)/3-(1.36943…)/3+(2pin)/3