What is the general solution for cos(2x)=1-2(3/5)^2 ?

The general solution for cos(2x)=1-2(3/5)^2 is x=(1.28700…)/2+pin,x=pi-(1.28700…)/2+pin

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cos(2x)=1-2(3/5)^2

cos (2 x) = 1 - 2 (\frac{3}{5})^{2}

x = \frac{1.28700 \dots}{2} + πn, x = π - \frac{1.28700 \dots}{2} + πn

Solution steps

cos (2 x) = 1 - 2 (\frac{3}{5})^{2}

Simplify

1 - 2 (\frac{3}{5})^{2} : \frac{7}{25}

Apply trig inverse properties

2 x = arccos (\frac{7}{25}) + 2 πn, 2 x = 2 π - arccos (\frac{7}{25}) + 2 πn

Solve

2 x = arccos (\frac{7}{25}) + 2 πn : x = \frac{arccos ( \frac{7}{25} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{7}{25}) + 2 πn : x = π - \frac{arccos ( \frac{7}{25} )}{2} + πn

x = \frac{arccos ( \frac{7}{25} )}{2} + πn, x = π - \frac{arccos ( \frac{7}{25} )}{2} + πn

Show solutions in decimal form

x = \frac{1.28700 \dots}{2} + πn, x = π - \frac{1.28700 \dots}{2} + πn

sin (θ + \frac{π}{3}) = \frac{1}{2}

4 sin^{2} (x) = 5 - 4 cos (x), 0 \leq x < 2 π

sin (x) - 6 cos^{2} (x) + 4 = 0

cos (θ) = \frac{3}{4}

2 sec (4 x) - 2 = 0

What is the general solution for cos(2x)=1-2(3/5)^2 ?
The general solution for cos(2x)=1-2(3/5)^2 is x=(1.28700…)/2+pin,x=pi-(1.28700…)/2+pin