Question

Solve the equation on the interval (0,2pi). Write your answer in exact simplest form.

$cos6x-5sin3x+2=0$

Answer 1

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Answer:

  x ∈ {π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18}

Step-by-step explanation:

Using the trig identity ...

  cos(2x) = 1 -2sin²(x)

we can rewrite the equation as ...

  (1 -2sin²(3x)) -5sin(3x) +2 = 0

  2sin²(3x) +5sin(3x) -3 = 0 . . . . . standard form

  (sin(3x) +3)(2sin(3x) -1) = 0 . . . . factored form

There are no real values of x that will make the first factor zero. The second factor is zero when ...

  sin(3x) = 1/2

  3x = arcsin(1/2) +2nπ   or   (2n+1)π -arcsin(1/2)

  x = 1/3(π/6 +2nπ)   or   1/3((2n +1)π -π/6)

  x = (12n+1)π/18  or  (12n+5)π/18

  x ∈ {π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18}