What is the general solution for 2(cos(x)+3)=5+cos(x) ?

The general solution for 2(cos(x)+3)=5+cos(x) is x=pi+2pin

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2(cos(x)+3)=5+cos(x)

2 (cos (x) + 3) = 5 + cos (x)

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 (cos (x) + 3) = 5 + cos (x)

Solve by substitution

2 (cos (x) + 3) = 5 + cos (x)

Let:

cos (x) = u

2 (u + 3) = 5 + u

2 (u + 3) = 5 + u : u = - 1

2 (u + 3) = 5 + u

Expand

2 (u + 3) : 2 u + 6

2 (u + 3)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = u, c = 3

= 2 u + 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 2 u + 6

2 u + 6 = 5 + u

Move

6

to the right side

2 u + 6 = 5 + u

Subtract

6

from both sides

2 u + 6 - 6 = 5 + u - 6

Simplify

2 u = u - 1

2 u = u - 1

Move

u

to the left side

2 u = u - 1

Subtract

u

from both sides

2 u - u = u - 1 - u

Simplify

u = - 1

u = - 1

Substitute back

u = cos (x)

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (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}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = π + 2 πn

4 cos (3 x) - 1 = 2 sin (3 x) tan (3 x)

tan (θ) = \frac{15}{20}

- cos (x) = 0.9

3 cos (θ) - 2 sin (θ) cos (θ) = 0

sec (x) + sin^{2} (x) + cos^{2} (x) = tan^{2} (x)

What is the general solution for 2(cos(x)+3)=5+cos(x) ?
The general solution for 2(cos(x)+3)=5+cos(x) is x=pi+2pin