Question

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.) 5 sin(2θ) − 2 sin(θ) = 0

Answer 1

Answer:

  x = {kπ, arccos(1/5) +2kπ, 2kπ -arccos(1/5)}

Step-by-step explanation:

The double-angle trig identity for sine is useful:

  5(2sin(θ)cos(θ)) -2sin(θ) = 0

  2sin(θ)(5cos(θ) -1) = 0

This has solutions that make the factors zero:

  θ = arcsin(0) = kπ

and ...

  cos(θ) = 1/5

  θ = arccos(1/5) +2kπ . . . . or . . . . 2kπ -arccos(1/5)

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Some numerical values are shown on the graph attached. values for multiples of pi are ...

  {..., -12.566, -9.425, -6.283, -3.142, 0, 3.142, 6.283, 9.425, 12.566, ...}