What is the general solution for 7cos(x)=5cos(x)+sqrt(3) ?

The general solution for 7cos(x)=5cos(x)+sqrt(3) is x= pi/6+2pin,x=(11pi)/6+2pin

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7cos(x)=5cos(x)+sqrt(3)

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

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

7 u = 5 u + 3

7 u = 5 u + 3 : u = \frac{3}{2}

7 u = 5 u + 3

Move

5 u

to the left side

7 u = 5 u + 3

Subtract

5 u

from both sides

7 u - 5 u = 5 u + 3 - 5 u

Simplify

2 u = 3

2 u = 3

Divide both sides by

2

2 u = 3

Divide both sides by

2

\frac{2 u}{2} = \frac{3}{2}

Simplify

u = \frac{3}{2}

u = \frac{3}{2}

Substitute back

u = cos (x)

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

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

cos (x) = \frac{3}{2} : x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

General solutions for

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

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}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

cos^{2} (a) - 3 sin^{2} (a) = 0

sin (α) = - \frac{3}{5}, π < α < \frac{3 π}{2}

cos^{4} (x) - sin^{4} (x) = 0

5 + 2 tan (3 θ + \frac{π}{3}) = 3, 0 \leq θ \leq π

sin (d) = \frac{12}{15}

What is the general solution for 7cos(x)=5cos(x)+sqrt(3) ?
The general solution for 7cos(x)=5cos(x)+sqrt(3) is x= pi/6+2pin,x=(11pi)/6+2pin