What is the general solution for 2cos^4(x)cos(x)-cos^5(x)=1 ?

The general solution for 2cos^4(x)cos(x)-cos^5(x)=1 is x=2pin

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2cos^4(x)cos(x)-cos^5(x)=1

2 cos^{4} (x) cos (x) - cos^{5} (x) = 1

x = 2 πn

Degrees

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

Solution steps

2 cos^{4} (x) cos (x) - cos^{5} (x) = 1

Simplify

2 cos^{4} (x) cos (x) - cos^{5} (x) : cos^{5} (x)

2 cos^{4} (x) cos (x) - cos^{5} (x)

2 cos^{4} (x) cos (x) = 2 cos^{5} (x)

2 cos^{4} (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

4 + 1 = 5

= 2 cos^{5} (x)

= 2 cos^{5} (x) - cos^{5} (x)

Add similar elements:

2 cos^{5} (x) - cos^{5} (x) = cos^{5} (x)

= cos^{5} (x)

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

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

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

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

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

\frac{sin ^{3} ( x )}{sin ( x )} = 0

What is the general solution for 2cos^4(x)cos(x)-cos^5(x)=1 ?
The general solution for 2cos^4(x)cos(x)-cos^5(x)=1 is x=2pin