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

The general solution for 5cos(x)+sqrt(2)=3cos(x) is x=(3pi)/4+2pin,x=(5pi)/4+2pin

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

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

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Degrees

x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

5 u + 2 = 3 u

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

5 u + 2 = 3 u

Move

2

to the right side

5 u + 2 = 3 u

Subtract

2

from both sides

5 u + 2 - 2 = 3 u - 2

Simplify

5 u = 3 u - 2

5 u = 3 u - 2

Move

3 u

to the left side

5 u = 3 u - 2

Subtract

3 u

from both sides

5 u - 3 u = 3 u - 2 - 3 u

Simplify

2 u = - 2

2 u = - 2

Divide both sides by

2

2 u = - 2

Divide both sides by

2

\frac{2 u}{2} = \frac{- 2}{2}

Simplify

u = - \frac{2}{2}

u = - \frac{2}{2}

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{2}{2} : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

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

General solutions for

cos (x) = - \frac{2}{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{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Combine all the solutions

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ)

cos (2 x) - cos^{2} (x) = 0

tan (π - θ) = tan (θ)

8 cos (x) = sec^{2} (x)

5 cot (θ) - 4 = 0

What is the general solution for 5cos(x)+sqrt(2)=3cos(x) ?
The general solution for 5cos(x)+sqrt(2)=3cos(x) is x=(3pi)/4+2pin,x=(5pi)/4+2pin