What is the general solution for (cos^2(a)-3cos(a)+2)/(sin^2(a))=1 ?

The general solution for (cos^2(a)-3cos(a)+2)/(sin^2(a))=1 is a= pi/3+2pin,a=(5pi)/3+2pin

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(cos^2(a)-3cos(a)+2)/(sin^2(a))=1

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

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

Solution steps

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

Subtract

1

from both sides

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 = 0

Simplify

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 : \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )}

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

1 + 2 cos^{2} (a) - 3 cos (a) = 0

Solve by substitution

cos (a) = 1, cos (a) = \frac{1}{2}

cos (a) = 1 : a = 2 πn

cos (a) = \frac{1}{2} : a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

2 πn

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

\frac{sin ( x ) - ( 2 )}{2} = 0

cos (2 x) = 5 - 6 cos^{2} (x)

cos^{4} (x) = 0.37

sin^{2} (x) = \frac{( 3 )}{2}

sin (x) cos (x) = 2 sin (2 x)

What is the general solution for (cos^2(a)-3cos(a)+2)/(sin^2(a))=1 ?
The general solution for (cos^2(a)-3cos(a)+2)/(sin^2(a))=1 is a= pi/3+2pin,a=(5pi)/3+2pin