What is the general solution for (cos(x)-4)(cos(x)+1)=0 ?

The general solution for (cos(x)-4)(cos(x)+1)=0 is x=pi+2pin

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(cos(x)-4)(cos(x)+1)=0

(cos (x) - 4) (cos (x) + 1) = 0

x = π + 2 πn

Degrees

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

Solution steps

(cos (x) - 4) (cos (x) + 1) = 0

Solving each part separately

cos (x) - 4 = 0 or cos (x) + 1 = 0

cos (x) - 4 = 0 :

No Solution

cos (x) - 4 = 0

Move

4

to the right side

cos (x) - 4 = 0

Add

4

to both sides

cos (x) - 4 + 4 = 0 + 4

Simplify

cos (x) = 4

cos (x) = 4

- 1 \leq cos (x) \leq 1

No Solution

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

cos (x) + 1 - 1 = 0 - 1

Simplify

cos (x) = - 1

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

- 15 + 4 cos (x) = - 15 - 22

tan^{2} (x) sin (x) = tan^{2} (x)

2 sin (θ) = 2, 0 \leq θ < 2 π

2 cos^{2} (t) - 3 cos (t) = - 1

cos (t) - cos (2 t) = - 0.75

What is the general solution for (cos(x)-4)(cos(x)+1)=0 ?
The general solution for (cos(x)-4)(cos(x)+1)=0 is x=pi+2pin