Problems with Trigonometric functions- how to solve?

Question

I aqm trying to solve this trigonometric question, but I am finding it rather difficult to solve:

Find the roots for sin 4x= - (1/√2), where 0 ≤ x ≤ 2π.

If you could also provide me with sort of a step-by-step guide, that would be really helpful since it may assist me in solving future questions.

Answer 1

From the geometry of a 45-45-90 triangle:

sin(pi / 4) = 1 / sqrt(2).

Negative sines occur in quadrants 3 and 4.

sin(x) = - 1 / sqrt(2) when x = 5pi / 4, 7pi / 4.

sin(4x) = - 1 / sqrt(2) has roots

4x = 5pi / 4 + 2n pi, 7pi / 4 + 2n pi.

Note that the roots can be increased by any multiple (positive or negative) of 2 pi.

Dividing by 4:

x = 5pi / 16 + n pi / 2, 7pi / 16 + n pi / 2.

Substituting n = 0, 1, 2, 3, gives the values in [0, 2pi]:

x = 5pi / 16, 7pi / 16, 13pi / 16, 15pi / 16, 21pi / 16, 23pi / 16, 29pi / 16, 31pi / 16.