What is the general solution for 8cos(2x)-1=2 ?

The general solution for 8cos(2x)-1=2 is x=(1.18639…)/2+pin,x=pi-(1.18639…)/2+pin

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8cos(2x)-1=2

8 cos (2 x) - 1 = 2

x = \frac{1.18639 \dots}{2} + πn, x = π - \frac{1.18639 \dots}{2} + πn

Solution steps

8 cos (2 x) - 1 = 2

Move

1

to the right side

8 cos (2 x) = 3

Divide both sides by

8

cos (2 x) = \frac{3}{8}

Apply trig inverse properties

2 x = arccos (\frac{3}{8}) + 2 πn, 2 x = 2 π - arccos (\frac{3}{8}) + 2 πn

Solve

2 x = arccos (\frac{3}{8}) + 2 πn : x = \frac{arccos ( \frac{3}{8} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{3}{8}) + 2 πn : x = π - \frac{arccos ( \frac{3}{8} )}{2} + πn

x = \frac{arccos ( \frac{3}{8} )}{2} + πn, x = π - \frac{arccos ( \frac{3}{8} )}{2} + πn

Show solutions in decimal form

x = \frac{1.18639 \dots}{2} + πn, x = π - \frac{1.18639 \dots}{2} + πn

2 cos (2 x) + 3 = 23

sin (x) cos (2 x) = \frac{1}{2} (1 + sin (x))

e^{s i n (x)} = 2

\frac{sec ( x )}{cot ( x ) tan ( x )} = sin (x)

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

What is the general solution for 8cos(2x)-1=2 ?
The general solution for 8cos(2x)-1=2 is x=(1.18639…)/2+pin,x=pi-(1.18639…)/2+pin