What is the general solution for (cos^{2022}(x))/((sin^{2022)(x)+cos^{2022}(x))}=1 ?

The general solution for (cos^{2022}(x))/((sin^{2022)(x)+cos^{2022}(x))}=1 is x=2pin,x=pi+2pin

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(cos^{2022}(x))/((sin^{2022)(x)+cos^{2022}(x))}=1

Solution

\frac{cos ^{2022} ( x )}{( sin ^{2022} ( x ) + cos ^{2022} ( x ) )} = 1

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

\frac{cos ^{2022} ( x )}{( sin ^{2022} ( x ) + cos ^{2022} ( x ) )} = 1

Subtract

1

from both sides

\frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - 1 = 0

Simplify

\frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - 1 : - \frac{sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

\frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - 1

Convert element to fraction:

1 = \frac{1 ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

= \frac{cos ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} - \frac{1 \cdot ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ^{2022} ( x ) - 1 \cdot ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

Multiply:

1 \cdot (sin^{2022} (x) + cos^{2022} (x)) = (sin^{2022} (x) + cos^{2022} (x))

= \frac{cos ^{2022} ( x ) - ( sin ^{2022} ( x ) + cos ^{2022} ( x ) )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

Expand

cos^{2022} (x) - (sin^{2022} (x) + cos^{2022} (x)) : - sin^{2022} (x)

cos^{2022} (x) - (sin^{2022} (x) + cos^{2022} (x))

- (sin^{2022} (x) + cos^{2022} (x)) : - sin^{2022} (x) - cos^{2022} (x)

- (sin^{2022} (x) + cos^{2022} (x))

Distribute parentheses

= - (sin^{2022} (x)) - (cos^{2022} (x))

Apply minus-plus rules

+ (- a) = - a

= - sin^{2022} (x) - cos^{2022} (x)

= cos^{2022} (x) - sin^{2022} (x) - cos^{2022} (x)

Add similar elements:

cos^{2022} (x) - cos^{2022} (x) = 0

= - sin^{2022} (x)

= \frac{- sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

Apply the fraction rule:

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

= - \frac{sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )}

- \frac{sin ^{2022} ( x )}{sin ^{2022} ( x ) + cos ^{2022} ( x )} = 0

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

- (sin^{2022} (x)) = 0

Divide both sides by

- 1

- (sin^{2022} (x)) = 0

Divide both sides by

- 1

\frac{- sin ^{2022} ( x )}{- 1} = \frac{0}{- 1}

Simplify

sin^{2022} (x) = 0

sin^{2022} (x) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

sin (x) = 0

The solution is

sin (x) = 0

General solutions for

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (cos^{2022}(x))/((sin^{2022)(x)+cos^{2022}(x))}=1 ?
The general solution for (cos^{2022}(x))/((sin^{2022)(x)+cos^{2022}(x))}=1 is x=2pin,x=pi+2pin