What is the general solution for 15cos(x)-5=4cos(x) ?

The general solution for 15cos(x)-5=4cos(x) is x=1.09893…+2pin,x=2pi-1.09893…+2pin

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15cos(x)-5=4cos(x)

15 cos (x) - 5 = 4 cos (x)

x = 1.09893 \dots + 2 πn, x = 2 π - 1.09893 \dots + 2 πn

Degrees

x = 62.96430 \dots^{\circ} + 36 0^{\circ} n, x = 297.03569 \dots^{\circ} + 36 0^{\circ} n

Solution steps

15 cos (x) - 5 = 4 cos (x)

Solve by substitution

15 cos (x) - 5 = 4 cos (x)

Let:

cos (x) = u

15 u - 5 = 4 u

15 u - 5 = 4 u : u = \frac{5}{11}

15 u - 5 = 4 u

Move

5

to the right side

15 u - 5 = 4 u

Add

5

to both sides

15 u - 5 + 5 = 4 u + 5

Simplify

15 u = 4 u + 5

15 u = 4 u + 5

Move

4 u

to the left side

15 u = 4 u + 5

Subtract

4 u

from both sides

15 u - 4 u = 4 u + 5 - 4 u

Simplify

11 u = 5

11 u = 5

Divide both sides by

11

11 u = 5

Divide both sides by

11

\frac{11 u}{11} = \frac{5}{11}

Simplify

u = \frac{5}{11}

u = \frac{5}{11}

Substitute back

u = cos (x)

cos (x) = \frac{5}{11}

cos (x) = \frac{5}{11}

cos (x) = \frac{5}{11} : x = arccos (\frac{5}{11}) + 2 πn, x = 2 π - arccos (\frac{5}{11}) + 2 πn

cos (x) = \frac{5}{11}

Apply trig inverse properties

cos (x) = \frac{5}{11}

General solutions for

cos (x) = \frac{5}{11}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{5}{11}) + 2 πn, x = 2 π - arccos (\frac{5}{11}) + 2 πn

x = arccos (\frac{5}{11}) + 2 πn, x = 2 π - arccos (\frac{5}{11}) + 2 πn

Combine all the solutions

x = arccos (\frac{5}{11}) + 2 πn, x = 2 π - arccos (\frac{5}{11}) + 2 πn

Show solutions in decimal form

x = 1.09893 \dots + 2 πn, x = 2 π - 1.09893 \dots + 2 πn

sin (θ) + cos (θ) = \frac{2 + 3}{2}

x, arctan (\frac{1}{x}) = 8 0^{\circ}

2 sin (θ) cos (θ) + 2 sin (θ) + cos (θ) + 1 = 0

cos (x) - cos (2 x) = 0, 0 \leq x \leq 2 π

sin (t) + 2 sin (2 t) = 0

What is the general solution for 15cos(x)-5=4cos(x) ?
The general solution for 15cos(x)-5=4cos(x) is x=1.09893…+2pin,x=2pi-1.09893…+2pin