What is the general solution for 5^{sin(2x-x/4)}=1 ?

The general solution for 5^{sin(2x-x/4)}=1 is x=(8pin)/7 ,x=(4pi)/7+(8pin)/7

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

5^{s i n (2 x - \frac{x}{4})} = 1

x = \frac{8 πn}{7}, x = \frac{4 π}{7} + \frac{8 πn}{7}

Solution steps

5^{s i n (2 x - \frac{x}{4})} = 1

Apply exponent rules

sin (2 x - \frac{x}{4}) ln (5) = ln (1)

Solve

sin (2 x - \frac{x}{4}) ln (5) = ln (1) : sin (2 x - \frac{x}{4}) = 0

sin (2 x - \frac{x}{4}) = 0

General solutions for

sin (2 x - \frac{x}{4}) = 0

2 x - \frac{x}{4} = 0 + 2 πn, 2 x - \frac{x}{4} = π + 2 πn

Solve

2 x - \frac{x}{4} = 0 + 2 πn : x = \frac{8 πn}{7}

Solve

2 x - \frac{x}{4} = π + 2 πn : x = \frac{4 π}{7} + \frac{8 πn}{7}

x = \frac{8 πn}{7}, x = \frac{4 π}{7} + \frac{8 πn}{7}

5 + tan (x) = 4

sin^{2} (x) - 6 sin (x) = 0

cos (a) = \frac{3}{3}

3 csc^{2} (x) - 2 csc (x) - 5 = 0

cos (2 x - 1) = 0.4

What is the general solution for 5^{sin(2x-x/4)}=1 ?
The general solution for 5^{sin(2x-x/4)}=1 is x=(8pin)/7 ,x=(4pi)/7+(8pin)/7