Question

find all solutions to each equation in radians sin8x-sin4x=0

Answer 1

Answer:

kπ/4, π/12 + nπ/2 and 5π/12 + nπ/2, where k, n are intergers.

Step-by-step explanation:

=> sin8x - sin4x = 0

=> (2sin4x.cos4x) - sin4x = 0

=> sin4x (2cos4x - 1) = 0

Either sin4x or 2cos4x - 1 is 0.

If sin4x = 0 : => sin4x = sin0 = sinπ

In general, sin4x = 0 = sinkπ

Therefore, x = kπ/4

If 2cos4x - 1 = 0 : => cos4x = 1/2

In general, cos4x = cos(π/3) = cos(5π/3)

Therefore, x = 1/4 (π/3) = 1/4 (5π/3)

As its value are repeated in period of 2π.

x = 1/4(π/3 + 2πn) and x = 1/4 (5π/3 + 2πn).

x = π/12 + nπ/2 and x = 5π/12 + nπ/2.

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