What is the general solution for tan^5(x)tan^2(x)=1 ?

The general solution for tan^5(x)tan^2(x)=1 is x= pi/4+pin

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tan^5(x)tan^2(x)=1

tan^{5} (x) tan^{2} (x) = 1

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{5} (x) tan^{2} (x) = 1

Simplify

tan^{5} (x) tan^{2} (x) : tan^{7} (x)

tan^{5} (x) tan^{2} (x)

Apply exponent rule:

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

tan^{5} (x) tan^{2} (x) = tan^{5 + 2} (x)

= tan^{5 + 2} (x)

Add the numbers:

5 + 2 = 7

= tan^{7} (x)

tan^{7} (x) = 1

For

x^{n} = f (a)

, n is odd, the solution is

Apply rule

= 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

tan (a) = \frac{5}{9}, b = 6

sin^{3} (x) sin^{2} (x) sin (x) = 0

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

sin^{2} (x) cos^{2} (x) = \frac{( 1 - cos ^{4} ( x ) )}{8}

What is the general solution for tan^5(x)tan^2(x)=1 ?
The general solution for tan^5(x)tan^2(x)=1 is x= pi/4+pin