What is the general solution for cos(1/2 x)=-1,y=2 ?

The general solution for cos(1/2 x)=-1,y=2 is No Solution for x\in\mathbb{R}

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cos(1/2 x)=-1,y=2

cos (\frac{1}{2} x) = - 1, y = 2

No Solution for x \in R

Solution steps

cos (\frac{1}{2} x) = - 1, y = 2

General solutions for

cos (\frac{1}{2} 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}

\frac{1}{2} x = π + 2 πn

\frac{1}{2} x = π + 2 πn

Solve

\frac{1}{2} x = π + 2 πn : x = 2 π + 4 πn

\frac{1}{2} x = π + 2 πn

Multiply both sides by

2

\frac{1}{2} x = π + 2 πn

Multiply both sides by

2

2 \cdot \frac{1}{2} x = 2 π + 2 \cdot 2 πn

Simplify

x = 2 π + 4 πn

x = 2 π + 4 πn

x = 2 π + 4 πn

Solutions for the range

y = 2

No Solution for x \in R

sin^{2} (a) = 2 sin^{2} (a) + 3

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

x, y = 2 sec (x)

- sin (2 x) - sin (x) = 0, 0 \leq x < 2 π

2 sin (6 x) sin (3 x) = 1 - cos (9 x)

What is the general solution for cos(1/2 x)=-1,y=2 ?
The general solution for cos(1/2 x)=-1,y=2 is No Solution for x\in\mathbb{R}