What is the general solution for cos(3x-pi/4)= 1/(sqrt(2)),0<= x<= pi ?

The general solution for cos(3x-pi/4)= 1/(sqrt(2)),0<= x<= pi is x= pi/6 ,x=(2pi)/3 ,x=(5pi)/6 ,x=0

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cos(3x-pi/4)= 1/(sqrt(2)),0<= x<= pi

cos (3 x - \frac{π}{4}) = \frac{1}{2}, 0 \leq x \leq π

x = \frac{π}{6}, x = \frac{2 π}{3}, x = \frac{5 π}{6}, x = 0

Solution steps

cos (3 x - \frac{π}{4}) = \frac{1}{2}, 0 \leq x \leq π

Simplify

\frac{1}{2} : \frac{2}{2}

General solutions for

cos (3 x - \frac{π}{4}) = \frac{2}{2}

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

Solve

3 x - \frac{π}{4} = \frac{π}{4} + 2 πn : x = \frac{π}{6} + \frac{2 πn}{3}

Solve

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

x = \frac{π}{6} + \frac{2 πn}{3}, x = \frac{2 πn}{3} + \frac{2 π}{3}

Solutions for the range

0 \leq x \leq π

x = \frac{π}{6}, x = \frac{2 π}{3}, x = \frac{5 π}{6}, x = 0

4 sin^{2} (x) = 3, 0 \leq x \leq 10 8^{\circ}

3 sin (x) + 3 cos (x) = 0

4 sin (2 x) + 4 cos (x) = 0

2 + cos (2 x) + 3 cos (x) = 0

2 cos (x) = tan (x) + sec (x)

What is the general solution for cos(3x-pi/4)= 1/(sqrt(2)),0<= x<= pi ?
The general solution for cos(3x-pi/4)= 1/(sqrt(2)),0<= x<= pi is x= pi/6 ,x=(2pi)/3 ,x=(5pi)/6 ,x=0