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

The general solution for sin^2(x)+cos^4(x)=n is

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

Solution

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

Solution

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

Solution steps

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

Subtract

n

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

1 - u^{2} + u^{4} - n = 0 : u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}, u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

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

Write in the standard form

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

u^{4} - u^{2} + 1 - n = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} - v + 1 - n = 0

Solve

v^{2} - v + 1 - n = 0 : v = \frac{1 + 4 n - 3}{2}, v = \frac{1 - 4 n - 3}{2}

v^{2} - v + 1 - n = 0

Solve with the quadratic formula

v^{2} - v + 1 - n = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 1 - n

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

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

Simplify

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

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

(- 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 (1 - n) = 4 (1 - n)

4 \cdot 1 \cdot (1 - n)

Multiply the numbers:

4 \cdot 1 = 4

= 4 (- n + 1)

= 1 - 4 (- n + 1)

Expand

1 - 4 (1 - n) : 4 n - 3

1 - 4 (1 - n)

Expand

- 4 (1 - n) : - 4 + 4 n

- 4 (1 - n)

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = n

= - 4 \cdot 1 - (- 4) n

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 n

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 n

= 1 - 4 + 4 n

Subtract the numbers:

1 - 4 = - 3

= 4 n - 3

= 4 n - 3

v_{1, 2} = \frac{- ( - 1 ) \pm 4 n - 3}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 4 n - 3}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 4 n - 3}{2 \cdot 1}

v = \frac{- ( - 1 ) + 4 n - 3}{2 \cdot 1} : \frac{1 + 4 n - 3}{2}

\frac{- ( - 1 ) + 4 n - 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 + 4 n - 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 4 n - 3}{2}

v = \frac{- ( - 1 ) - 4 n - 3}{2 \cdot 1} : \frac{1 - 4 n - 3}{2}

\frac{- ( - 1 ) - 4 n - 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 - 4 n - 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 4 n - 3}{2}

The solutions to the quadratic equation are:

v = \frac{1 + 4 n - 3}{2}, v = \frac{1 - 4 n - 3}{2}

v = \frac{1 + 4 n - 3}{2}, v = \frac{1 - 4 n - 3}{2}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = \frac{1 + 4 n - 3}{2} : u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}

u^{2} = \frac{1 + 4 n - 3}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}

Solve

u^{2} = \frac{1 - 4 n - 3}{2} : u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

u^{2} = \frac{1 - 4 n - 3}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

The solutions are

u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}, u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 4 n - 3}{2}, cos (x) = - \frac{1 + 4 n - 3}{2}, cos (x) = \frac{1 - 4 n - 3}{2}, cos (x) = - \frac{1 - 4 n - 3}{2}

cos (x) = \frac{1 + 4 n - 3}{2}, cos (x) = - \frac{1 + 4 n - 3}{2}, cos (x) = \frac{1 - 4 n - 3}{2}, cos (x) = - \frac{1 - 4 n - 3}{2}

cos (x) = \frac{1 + 4 n - 3}{2} : x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = \frac{1 + 4 n - 3}{2}

Apply trig inverse properties

cos (x) = \frac{1 + 4 n - 3}{2}

General solutions for

cos (x) = \frac{1 + 4 n - 3}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πk, x = - arccos (a) + 2 πk

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 + 4 n - 3}{2} : x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 + 4 n - 3}{2}

Apply trig inverse properties

cos (x) = - \frac{1 + 4 n - 3}{2}

General solutions for

cos (x) = - \frac{1 + 4 n - 3}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πk, x = - arccos (a) + 2 πk

x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk

x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = \frac{1 - 4 n - 3}{2} : x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk

cos (x) = \frac{1 - 4 n - 3}{2}

Apply trig inverse properties

cos (x) = \frac{1 - 4 n - 3}{2}

General solutions for

cos (x) = \frac{1 - 4 n - 3}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πk, x = - arccos (a) + 2 πk

x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk

x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 - 4 n - 3}{2} : x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 - 4 n - 3}{2}

Apply trig inverse properties

cos (x) = - \frac{1 - 4 n - 3}{2}

General solutions for

cos (x) = - \frac{1 - 4 n - 3}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πk, x = - arccos (a) + 2 πk

x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

Combine all the solutions

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

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Frequently Asked Questions (FAQ)

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