Question

Sometimes a trigonometric equation is written in quadratic form. This means that you can solve the equation the same way you would solve a standard quadratic equation. For example, the equation: sin2 2.5x − 4 sin 2.5x − 5 = 0 can be considered to be like the form x 2 − 4x − 5 = 0. Use this fact to solve this trigonometric equation on the interval 0 ≤ x < 2π.

Answer 1

Answer:

108 degrees

Step-by-step explanation:

To solve this adequately, we simply make a substitution. We can say let sin2.5x = x

Hence we can thus have the quadratic equation form which we can solve.

x ^2 − 4x − 5 = 0.

Solving this yields the following:

x^2 +x - 5x -5 = 0

x( x + 1) -5(x + 1) = 0

(x - 5) (x + 1) = 0

x = 5 or -1

Recall the substitution:

sin2.5x = x

We cannot use the value -5 as the value of the sine function cannot be in this range. We thus ignore it and pick the -1 answer only.

Sin2.5x = -1

2.5x = arcsin(-1) = 270

2.5x = 270

x = 270/2.5 = 108 degrees