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

Solution

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

Solution

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

Degrees

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

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 )} - 1

Convert element to fraction:

1 = \frac{1 sin ^{2} ( a )}{sin ^{2} ( a )}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Multiply:

1 \cdot sin^{2} (a) = sin^{2} (a)

= \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

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

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

Simplify

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

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

- (1 - cos^{2} (a)) : - 1 + cos^{2} (a)

- (1 - cos^{2} (a))

Distribute parentheses

= - (1) - (- cos^{2} (a))

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

cos^{2} (a) + cos^{2} (a) = 2 cos^{2} (a)

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

Add/Subtract the numbers:

2 - 1 = 1

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

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

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

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

Solve by substitution

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

Let:

cos (a) = u

1 + 2 u^{2} - 3 u = 0

1 + 2 u^{2} - 3 u = 0 : u = 1, u = \frac{1}{2}

1 + 2 u^{2} - 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

(- 3)^{2} - 4 \cdot 2 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 2}

u = \frac{- ( - 3 ) + 1}{2 \cdot 2} : 1

\frac{- ( - 3 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

Add the numbers:

3 + 1 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

Subtract the numbers:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{2}

Substitute back

u = cos (a)

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

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

cos (a) = 1 : a = 2 πn

cos (a) = 1

General solutions for

cos (a) = 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}

a = 0 + 2 πn

a = 0 + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

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

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

General solutions for

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

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

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

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Frequently Asked Questions (FAQ)

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