What is the general solution for 7cos(4x)=6 ?

The general solution for 7cos(4x)=6 is x=(0.54109…)/4+(pin)/2 ,x= pi/2-(0.54109…)/4+(pin)/2

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7cos(4x)=6

7 cos (4 x) = 6

x = \frac{0.54109 \dots}{4} + \frac{πn}{2}, x = \frac{π}{2} - \frac{0.54109 \dots}{4} + \frac{πn}{2}

Solution steps

7 cos (4 x) = 6

Divide both sides by

7

cos (4 x) = \frac{6}{7}

Apply trig inverse properties

4 x = arccos (\frac{6}{7}) + 2 πn, 4 x = 2 π - arccos (\frac{6}{7}) + 2 πn

Solve

4 x = arccos (\frac{6}{7}) + 2 πn : x = \frac{arccos ( \frac{6}{7} )}{4} + \frac{πn}{2}

Solve

4 x = 2 π - arccos (\frac{6}{7}) + 2 πn : x = \frac{π}{2} - \frac{arccos ( \frac{6}{7} )}{4} + \frac{πn}{2}

x = \frac{arccos ( \frac{6}{7} )}{4} + \frac{πn}{2}, x = \frac{π}{2} - \frac{arccos ( \frac{6}{7} )}{4} + \frac{πn}{2}

Show solutions in decimal form

x = \frac{0.54109 \dots}{4} + \frac{πn}{2}, x = \frac{π}{2} - \frac{0.54109 \dots}{4} + \frac{πn}{2}

sin (x) tan (x) + 3 tan (x) = 0

sin^{2} (\frac{x}{2}) = 0

tan (t) = 2

- 120 = - 90 - arctan (0.1) + arctan (0.1 x)

3 tan (x) = 0

What is the general solution for 7cos(4x)=6 ?
The general solution for 7cos(4x)=6 is x=(0.54109…)/4+(pin)/2 ,x= pi/2-(0.54109…)/4+(pin)/2