What is the general solution for 7cos(x)+12=6cos(x)+13 ?

The general solution for 7cos(x)+12=6cos(x)+13 is x=2pin

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7cos(x)+12=6cos(x)+13

7 cos (x) + 12 = 6 cos (x) + 13

x = 2 πn

Degrees

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

Solution steps

7 cos (x) + 12 = 6 cos (x) + 13

Solve by substitution

7 cos (x) + 12 = 6 cos (x) + 13

Let:

cos (x) = u

7 u + 12 = 6 u + 13

7 u + 12 = 6 u + 13 : u = 1

7 u + 12 = 6 u + 13

Move

12

to the right side

7 u + 12 = 6 u + 13

Subtract

12

from both sides

7 u + 12 - 12 = 6 u + 13 - 12

Simplify

7 u = 6 u + 1

7 u = 6 u + 1

Move

6 u

to the left side

7 u = 6 u + 1

Subtract

6 u

from both sides

7 u - 6 u = 6 u + 1 - 6 u

Simplify

u = 1

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

cos (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = 2 πn

sin (\frac{x}{2}) + cos (\frac{x}{2}) = 2

cos (θ) = - \frac{1}{2}, 0 \leq θ < 2 π

2 tan (x) - 16 = 0, 0 \leq x < \frac{π}{2}

x, sin (x) = \frac{2}{2}

4 sec (x) + 6 = - 2

What is the general solution for 7cos(x)+12=6cos(x)+13 ?
The general solution for 7cos(x)+12=6cos(x)+13 is x=2pin