What is the general solution for cos(2x)=0.98 ?

The general solution for cos(2x)=0.98 is x=(0.20033…)/2+pin,x=pi-(0.20033…)/2+pin

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cos(2x)=0.98

cos (2 x) = 0.98

x = \frac{0.20033 \dots}{2} + πn, x = π - \frac{0.20033 \dots}{2} + πn

Degrees

x = 5.73917 \dots^{\circ} + 18 0^{\circ} n, x = 174.26082 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 x) = 0.98

Apply trig inverse properties

cos (2 x) = 0.98

General solutions for

cos (2 x) = 0.98

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 x = arccos (0.98) + 2 πn, 2 x = 2 π - arccos (0.98) + 2 πn

2 x = arccos (0.98) + 2 πn, 2 x = 2 π - arccos (0.98) + 2 πn

Solve

2 x = arccos (0.98) + 2 πn : x = \frac{arccos ( 0.98 )}{2} + πn

2 x = arccos (0.98) + 2 πn

Divide both sides by

2

2 x = arccos (0.98) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( 0.98 )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( 0.98 )}{2} + πn

x = \frac{arccos ( 0.98 )}{2} + πn

Solve

2 x = 2 π - arccos (0.98) + 2 πn : x = π - \frac{arccos ( 0.98 )}{2} + πn

2 x = 2 π - arccos (0.98) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (0.98) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( 0.98 )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arccos ( 0.98 )}{2} + πn

x = π - \frac{arccos ( 0.98 )}{2} + πn

x = \frac{arccos ( 0.98 )}{2} + πn, x = π - \frac{arccos ( 0.98 )}{2} + πn

Show solutions in decimal form

x = \frac{0.20033 \dots}{2} + πn, x = π - \frac{0.20033 \dots}{2} + πn

8 sin^{2} (x) + 2 sin (x) - 3 = 0

2 (arcsin (x))^{2} + (2 - π) arcsin (x) - π = 0

cos^{2} (x) + 3 sin (- x) - 2 = 0

cos (x) = - 1, 0 < x < 2 π

\frac{1}{1 + cos ( x )} + \frac{1}{1 - cos ( x )} = 32

What is the general solution for cos(2x)=0.98 ?
The general solution for cos(2x)=0.98 is x=(0.20033…)/2+pin,x=pi-(0.20033…)/2+pin