What is the general solution for 6cos(2x)+cos(2x)=1 ?

The general solution for 6cos(2x)+cos(2x)=1 is x=(1.42744…)/2+pin,x=pi-(1.42744…)/2+pin

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6cos(2x)+cos(2x)=1

6 cos (2 x) + cos (2 x) = 1

x = \frac{1.42744 \dots}{2} + πn, x = π - \frac{1.42744 \dots}{2} + πn

Solution steps

6 cos (2 x) + cos (2 x) = 1

Simplify

6 cos (2 x) + cos (2 x) : 7 cos (2 x)

7 cos (2 x) = 1

Divide both sides by

7

cos (2 x) = \frac{1}{7}

Apply trig inverse properties

2 x = arccos (\frac{1}{7}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{7}) + 2 πn

Solve

2 x = arccos (\frac{1}{7}) + 2 πn : x = \frac{arccos ( \frac{1}{7} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{1}{7}) + 2 πn : x = π - \frac{arccos ( \frac{1}{7} )}{2} + πn

x = \frac{arccos ( \frac{1}{7} )}{2} + πn, x = π - \frac{arccos ( \frac{1}{7} )}{2} + πn

Show solutions in decimal form

x = \frac{1.42744 \dots}{2} + πn, x = π - \frac{1.42744 \dots}{2} + πn

0.8 = \frac{1}{2} (1 - cos (\frac{θ}{2}))

2 tan^{2} (x) + tan (x) - 2 = 0

3 sec (θ) + 1 = 7, 0 \leq θ < 2 π

sin (A) cos (A) = sin (A)

tan (\frac{x}{2} - \frac{π}{4}) = 1

What is the general solution for 6cos(2x)+cos(2x)=1 ?
The general solution for 6cos(2x)+cos(2x)=1 is x=(1.42744…)/2+pin,x=pi-(1.42744…)/2+pin