What is the general solution for 2cos(4x)-sqrt(2)=0 ?

The general solution for 2cos(4x)-sqrt(2)=0 is x= pi/(16)+(pin)/2 ,x=(7pi}{16}+\frac{pin)/2

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2cos(4x)-sqrt(2)=0

2 cos (4 x) - 2 = 0

x = \frac{π}{16} + \frac{πn}{2}, x = \frac{7 π}{16} + \frac{πn}{2}

Solution steps

2 cos (4 x) - 2 = 0

Move

2

to the right side

2 cos (4 x) = 2

Divide both sides by

2

cos (4 x) = \frac{2}{2}

General solutions for

cos (4 x) = \frac{2}{2}

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

Solve

4 x = \frac{π}{4} + 2 πn : x = \frac{π}{16} + \frac{πn}{2}

Solve

4 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}, x = \frac{7 π}{16} + \frac{πn}{2}

\frac{3}{2} tan (2 x + 135) = - \frac{3}{2}

6 sin^{2} (θ) - 5 sin (θ) = - 1

csc (x) = - 4

tan (x) = - 3, - 27 0^{\circ} \leq x \leq 27 0^{\circ}

5 cos^{2} (θ) = - 13 cos (θ) + 6

What is the general solution for 2cos(4x)-sqrt(2)=0 ?
The general solution for 2cos(4x)-sqrt(2)=0 is x= pi/(16)+(pin)/2 ,x=(7pi}{16}+\frac{pin)/2