What is the general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi ?

The general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi is a= pi/8 ,a=(7pi)/8 ,a=(9pi)/8 ,a=(15pi)/8

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2cos(2a)=sqrt(2),0<= θ<= 2pi

2 cos (2 a) = 2, 0 \leq θ \leq 2 π

a = \frac{π}{8}, a = \frac{7 π}{8}, a = \frac{9 π}{8}, a = \frac{15 π}{8}

Solution steps

2 cos (2 a) = 2, 0 \leq θ \leq 2 π

Divide both sides by

2

cos (2 a) = \frac{2}{2}

General solutions for

cos (2 a) = \frac{2}{2}

2 a = \frac{π}{4} + 2 πn, 2 a = \frac{7 π}{4} + 2 πn

Solve

2 a = \frac{π}{4} + 2 πn : a = \frac{π}{8} + πn

Solve

2 a = \frac{7 π}{4} + 2 πn : a = \frac{7 π}{8} + πn

a = \frac{π}{8} + πn, a = \frac{7 π}{8} + πn

Solutions for the range

0 \leq θ \leq 2 π

a = \frac{π}{8}, a = \frac{7 π}{8}, a = \frac{9 π}{8}, a = \frac{15 π}{8}

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tan (x) = 2 - 1

What is the general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi ?
The general solution for 2cos(2a)=sqrt(2),0<= θ<= 2pi is a= pi/8 ,a=(7pi)/8 ,a=(9pi)/8 ,a=(15pi)/8