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

The general solution for 2cos(x)=-cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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2cos(x)=-cos(x)

2 cos (x) = - cos (x)

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) = - cos (x)

Solve by substitution

2 cos (x) = - cos (x)

Let:

cos (x) = u

2 u = - u

2 u = - u : u = 0

2 u = - u

Move

u

to the left side

2 u = - u

Add

u

to both sides

2 u + u = - u + u

Simplify

3 u = 0

3 u = 0

Divide both sides by

3

3 u = 0

Divide both sides by

3

\frac{3 u}{3} = \frac{0}{3}

Simplify

u = 0

u = 0

Substitute back

u = cos (x)

cos (x) = 0

cos (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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}

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

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

Combine all the solutions

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

tan (x) = - 10

cos (\frac{π}{x}) = \frac{3}{3}

y, arctan (\frac{y}{x}) = ln (x)

- cos (7 4^{\circ}) = cos (θ)

sin (x) = 50

What is the general solution for 2cos(x)=-cos(x) ?
The general solution for 2cos(x)=-cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin