What is the general solution for cos(2x-pi/4)=0,0<= x<= 2pi ?

The general solution for cos(2x-pi/4)=0,0<= x<= 2pi is x=(3pi)/8 ,x=(7pi)/8 ,x=(11pi)/8 ,x=(15pi)/8

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cos(2x-pi/4)=0,0<= x<= 2pi

cos (2 x - \frac{π}{4}) = 0, 0 \leq x \leq 2 π

x = \frac{3 π}{8}, x = \frac{7 π}{8}, x = \frac{11 π}{8}, x = \frac{15 π}{8}

Solution steps

cos (2 x - \frac{π}{4}) = 0, 0 \leq x \leq 2 π

General solutions for

cos (2 x - \frac{π}{4}) = 0

2 x - \frac{π}{4} = \frac{π}{2} + 2 πn, 2 x - \frac{π}{4} = \frac{3 π}{2} + 2 πn

Solve

2 x - \frac{π}{4} = \frac{π}{2} + 2 πn : x = πn + \frac{3 π}{8}

Solve

2 x - \frac{π}{4} = \frac{3 π}{2} + 2 πn : x = πn + \frac{7 π}{8}

x = πn + \frac{3 π}{8}, x = πn + \frac{7 π}{8}

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{8}, x = \frac{7 π}{8}, x = \frac{11 π}{8}, x = \frac{15 π}{8}

8 - sin (θ) = cos (2 θ)

x, 2 cos^{2} (x) - cos (x) - 1 = 0

- 8 cos (- 4 θ + 150) = 43

4 cos^{2} (x) + 2 sin (x) - 1 = 0

cos (x) = \frac{- 3}{2}, 0 \leq x \leq 2 π

What is the general solution for cos(2x-pi/4)=0,0<= x<= 2pi ?
The general solution for cos(2x-pi/4)=0,0<= x<= 2pi is x=(3pi)/8 ,x=(7pi)/8 ,x=(11pi)/8 ,x=(15pi)/8