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

The general solution for 1+cos^2(x)=sin^4(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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1+cos^2(x)=sin^4(x)

Solution

1 + cos^{2} (x) = sin^{4} (x)

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + cos^{2} (x) = sin^{4} (x)

Subtract

sin^{4} (x)

from both sides

1 + cos^{2} (x) - sin^{4} (x) = 0

Rewrite using trig identities

1 + cos^{2} (x) - sin^{4} (x)

Use the Pythagorean identity:

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

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

= 1 + 1 - sin^{2} (x) - sin^{4} (x)

Simplify

= - sin^{4} (x) - sin^{2} (x) + 2

2 - sin^{2} (x) - sin^{4} (x) = 0

Solve by substitution

2 - sin^{2} (x) - sin^{4} (x) = 0

Let:

sin (x) = u

2 - u^{2} - u^{4} = 0

2 - u^{2} - u^{4} = 0 : u = 2 i, u = - 2 i, u = 1, u = - 1

2 - u^{2} - u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{4} - u^{2} + 2 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

- v^{2} - v + 2 = 0

Solve

- v^{2} - v + 2 = 0 : v = - 2, v = 1

- v^{2} - v + 2 = 0

Solve with the quadratic formula

- v^{2} - v + 2 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

1 + 3 = 4

= \frac{4}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{- 2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

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

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

Remove parentheses:

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

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

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

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

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

v = - 2, v = 1

v = - 2, v = 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = - 2 : u = 2 i, u = - 2 i

u^{2} = - 2

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 2, u = - - 2

Simplify

- 2 : 2 i

- 2

Apply radical rule:

- a = - 1 a

- 2 = - 1 2

= - 1 2

Apply imaginary number rule:

- 1 = i

= 2 i

Simplify

- - 2 : - 2 i

- - 2

Simplify

- 2 : 2 i

- 2

Apply radical rule:

- a = - 1 a

- 2 = - 1 2

= - 1 2

Apply imaginary number rule:

- 1 = i

= 2 i

= - 2 i

u = 2 i, u = - 2 i

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 2 i, u = - 2 i, u = 1, u = - 1

Substitute back

u = sin (x)

sin (x) = 2 i, sin (x) = - 2 i, sin (x) = 1, sin (x) = - 1

sin (x) = 2 i, sin (x) = - 2 i, sin (x) = 1, sin (x) = - 1

sin (x) = 2 i :

No Solution

sin (x) = 2 i

No Solution

sin (x) = - 2 i :

No Solution

sin (x) = - 2 i

No Solution

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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arctan (x + 2) = arcsin (\frac{7}{25}) + arccos (\frac{4}{5})

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tan (c) = 0.6538

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Frequently Asked Questions (FAQ)

What is the general solution for 1+cos^2(x)=sin^4(x) ?
The general solution for 1+cos^2(x)=sin^4(x) is x= pi/2+2pin,x=(3pi)/2+2pin