What is the general solution for (cos^2(x)-4)/(cos(x)-2)=1 ?

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

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(cos^2(x)-4)/(cos(x)-2)=1

\frac{cos ^{2} ( x ) - 4}{cos ( x ) - 2} = 1

x = π + 2 πn

Degrees

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

Solution steps

\frac{cos ^{2} ( x ) - 4}{cos ( x ) - 2} = 1

Simplify

\frac{cos ^{2} ( x ) - 4}{cos ( x ) - 2} : cos (x) + 2

\frac{cos ^{2} ( x ) - 4}{cos ( x ) - 2}

Factor

cos^{2} (x) - 4 : (cos (x) + 2) (cos (x) - 2)

cos^{2} (x) - 4

Rewrite

4

2^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (x) + 2) (cos (x) - 2)

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

Cancel the common factor:

cos (x) - 2

= cos (x) + 2

cos (x) + 2 = 1

Move

2

to the right side

cos (x) + 2 = 1

Subtract

2

from both sides

cos (x) + 2 - 2 = 1 - 2

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

- 8 sec^{2} (x) - 26 tan (x) = 12

tan (θ) = 0.3

tan (θ) = 0.8

sin (a) tan (a) = sin (a)

2 tan (x) + sec^{2} (x) = 0

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