What is the general solution for 3cos(4x)=2 ?

The general solution for 3cos(4x)=2 is x=(0.84106…)/4+(pin)/2 ,x= pi/2-(0.84106…)/4+(pin)/2

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3cos(4x)=2

3 cos (4 x) = 2

x = \frac{0.84106 \dots}{4} + \frac{πn}{2}, x = \frac{π}{2} - \frac{0.84106 \dots}{4} + \frac{πn}{2}

Solution steps

3 cos (4 x) = 2

Divide both sides by

3

cos (4 x) = \frac{2}{3}

Apply trig inverse properties

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

Solve

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

Solve

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

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

Show solutions in decimal form

x = \frac{0.84106 \dots}{4} + \frac{πn}{2}, x = \frac{π}{2} - \frac{0.84106 \dots}{4} + \frac{πn}{2}

2 cos^{2} (x) = cos (x) + 1

4 cos (θ) + 2 = 0

6 cos^{2} (θ) = 3

cos (x) + cos (3 x) = 0

cos (x) = 3

What is the general solution for 3cos(4x)=2 ?
The general solution for 3cos(4x)=2 is x=(0.84106…)/4+(pin)/2 ,x= pi/2-(0.84106…)/4+(pin)/2