What is the general solution for cos(4*x)=-1 ?

The general solution for cos(4*x)=-1 is x= pi/4+(pin)/2

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cos(4*x)=-1

cos (4 \cdot x) = - 1

x = \frac{π}{4} + \frac{πn}{2}

Degrees

x = 4 5^{\circ} + 9 0^{\circ} n

Solution steps

cos (4 x) = - 1

General solutions for

cos (4 x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

4 x = π + 2 πn

4 x = π + 2 πn

Solve

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

4 x = π + 2 πn

Divide both sides by

4

4 x = π + 2 πn

Divide both sides by

4

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

Simplify

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

3 + 2 cos (x) = 4 cos (\frac{x}{2})

- 3 sin (x) = cos (x)

2 csc^{2} (x) = 5 - 5 cot (x)

sin (x + 0.5) = 0.92, 0 \leq x \leq 2 π

cos (4 x) = 0.8 - 0.6

What is the general solution for cos(4*x)=-1 ?
The general solution for cos(4*x)=-1 is x= pi/4+(pin)/2