What is the general solution for cos(2x)-0.25=0 ?

The general solution for cos(2x)-0.25=0 is x=(1.31811…)/2+pin,x=pi-(1.31811…)/2+pin

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cos(2x)-0.25=0

cos (2 x) - 0.25 = 0

x = \frac{1.31811 \dots}{2} + πn, x = π - \frac{1.31811 \dots}{2} + πn

Solution steps

cos (2 x) - 0.25 = 0

Move

0.25

to the right side

cos (2 x) = 0.25

Apply trig inverse properties

2 x = arccos (0.25) + 2 πn, 2 x = 2 π - arccos (0.25) + 2 πn

Solve

2 x = arccos (0.25) + 2 πn : x = \frac{arccos ( 0.25 )}{2} + πn

Solve

2 x = 2 π - arccos (0.25) + 2 πn : x = π - \frac{arccos ( 0.25 )}{2} + πn

x = \frac{arccos ( 0.25 )}{2} + πn, x = π - \frac{arccos ( 0.25 )}{2} + πn

Show solutions in decimal form

x = \frac{1.31811 \dots}{2} + πn, x = π - \frac{1.31811 \dots}{2} + πn

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

sec (x) = - \frac{4}{3}

sin (B) = \frac{1}{4}

55 sin (θ) - 20 - 20 cos (θ) = 0

y, sin (y) = 0

What is the general solution for cos(2x)-0.25=0 ?
The general solution for cos(2x)-0.25=0 is x=(1.31811…)/2+pin,x=pi-(1.31811…)/2+pin