What is the general solution for cos^2(x)+sin^2(x)=cos^5(x) ?

The general solution for cos^2(x)+sin^2(x)=cos^5(x) is x=2pin

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cos^2(x)+sin^2(x)=cos^5(x)

cos^{2} (x) + sin^{2} (x) = cos^{5} (x)

x = 2 πn

Degrees

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

Solution steps

cos^{2} (x) + sin^{2} (x) = cos^{5} (x)

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

1 = cos^{5} (x)

Switch sides

cos^{5} (x) = 1

For

x^{n} = f (a)

, n is odd, the solution is

Apply rule

= 1

cos (x) = 1

General solutions for

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 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

sin (x - 4 5^{5}) = \frac{( 2 )}{2}

\frac{sin ( x ) - 3 \cdot cos ( x )}{2} = 0

cos (\frac{1}{3 x}) = \frac{1}{3}

arctan (1 + x) + arctan (1 - x) = arctan (\frac{1}{2})

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

What is the general solution for cos^2(x)+sin^2(x)=cos^5(x) ?
The general solution for cos^2(x)+sin^2(x)=cos^5(x) is x=2pin