What is the general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) ?

The general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) is a=2pin,a=pi+2pin

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1-tan^4(a)cos^4(a)=1-2sin^2(a)

Solution

1 - tan^{4} (a) cos^{4} (a) = 1 - 2 sin^{2} (a)

Solution

a = 2 πn, a = π + 2 πn

Degrees

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

1 - tan^{4} (a) cos^{4} (a) = 1 - 2 sin^{2} (a)

Subtract

1 - 2 sin^{2} (a)

from both sides

2 sin^{2} (a) - tan^{4} (a) cos^{4} (a) = 0

Factor

2 sin^{2} (a) - tan^{4} (a) cos^{4} (a) : (2 sin (a) + tan^{2} (a) cos^{2} (a)) (2 sin (a) - tan^{2} (a) cos^{2} (a))

2 sin^{2} (a) - tan^{4} (a) cos^{4} (a)

Rewrite

tan^{4} (a) cos^{4} (a)

(tan (a) cos (a))^{4}

tan^{4} (a) cos^{4} (a)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

tan^{4} (a) cos^{4} (a) = (tan (a) cos (a))^{4}

= (tan (a) cos (a))^{4}

= 2 sin^{2} (a) - (tan (a) cos (a))^{4}

Rewrite

2 sin^{2} (a) - (tan (a) cos (a))^{4}

(2 sin (a))^{2} - ((tan (a) cos (a))^{2})^{2}

2 sin^{2} (a) - (tan (a) cos (a))^{4}

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} sin^{2} (a) - (tan (a) cos (a))^{4}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

(tan (a) cos (a))^{4} = ((tan (a) cos (a))^{2})^{2}

= (2)^{2} sin^{2} (a) - ((tan (a) cos (a))^{2})^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

(2)^{2} sin^{2} (a) = (2 sin (a))^{2}

= (2 sin (a))^{2} - ((tan (a) cos (a))^{2})^{2}

= (2 sin (a))^{2} - ((tan (a) cos (a))^{2})^{2}

Apply Difference of Two Squares Formula:

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

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

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

Simplify

(2 sin (a) + (tan (a) cos (a))^{2}) (2 sin (a) - (tan (a) cos (a))^{2}) : (2 sin (a) + tan^{2} (a) cos^{2} (a)) (2 sin (a) - tan^{2} (a) cos^{2} (a))

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

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

Solving each part separately

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

2 sin (a) + tan^{2} (a) cos^{2} (a) = 0 : a = 2 πn, a = π + 2 πn

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

Rewrite using trig identities

cos^{2} (a) tan^{2} (a) + sin (a) 2

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

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

cos^{2} (a) (\frac{sin ( a )}{cos ( a )})^{2}

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

(\frac{sin ( a )}{cos ( a )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (a)

= sin^{2} (a)

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

sin^{2} (a) + sin (a) 2 = 0

Solve by substitution

sin^{2} (a) + sin (a) 2 = 0

Let:

sin (a) = u

u^{2} + u 2 = 0

u^{2} + u 2 = 0 : u = 0, u = - 2

u^{2} + u 2 = 0

Write in the standard form

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

u^{2} + 2 u = 0

Solve with the quadratic formula

u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(2)^{2} - 4 \cdot 1 \cdot 0 = 2

(2)^{2} - 4 \cdot 1 \cdot 0

(2)^{2} = 2

(2)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 2 - 0

Subtract the numbers:

2 - 0 = 2

= 2

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

Separate the solutions

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

u = \frac{- 2 + 2}{2 \cdot 1} : 0

\frac{- 2 + 2}{2 \cdot 1}

Add similar elements:

- 2 + 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 2 - 2}{2 \cdot 1} : - 2

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

Add similar elements:

- 2 - 2 = - 22

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 2

The solutions to the quadratic equation are:

u = 0, u = - 2

Substitute back

u = sin (a)

sin (a) = 0, sin (a) = - 2

sin (a) = 0, sin (a) = - 2

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

General solutions for

sin (a) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 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, a = π + 2 πn

sin (a) = - 2 :

No Solution

sin (a) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

a = 2 πn, a = π + 2 πn

2 sin (a) - tan^{2} (a) cos^{2} (a) = 0 : a = 2 πn, a = π + 2 πn

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

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

cos^{2} (a) (\frac{sin ( a )}{cos ( a )})^{2}

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

(\frac{sin ( a )}{cos ( a )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (a)

= sin^{2} (a)

= - sin^{2} (a) + 2 sin (a)

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

Solve by substitution

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

Let:

sin (a) = u

- u^{2} + u 2 = 0

- u^{2} + u 2 = 0 : u = 0, u = 2

- u^{2} + u 2 = 0

Write in the standard form

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

- u^{2} + 2 u = 0

Solve with the quadratic formula

- u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = - 1, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

(2)^{2} - 4 (- 1) \cdot 0 = 2

(2)^{2} - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= (2)^{2} + 4 \cdot 1 \cdot 0

(2)^{2} = 2

(2)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 2 + 0

Add the numbers:

2 + 0 = 2

= 2

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

Separate the solutions

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

u = \frac{- 2 + 2}{2 ( - 1 )} : 0

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

Remove parentheses:

(- a) = - a

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

Add similar elements:

- 2 + 2 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

Remove parentheses:

(- a) = - a

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

Add similar elements:

- 2 - 2 = - 22

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2

The solutions to the quadratic equation are:

u = 0, u = 2

Substitute back

u = sin (a)

sin (a) = 0, sin (a) = 2

sin (a) = 0, sin (a) = 2

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

General solutions for

sin (a) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 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, a = π + 2 πn

sin (a) = 2 :

No Solution

sin (a) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

a = 2 πn, a = π + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) ?
The general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) is a=2pin,a=pi+2pin