Finding All Solutions of the Equation 10sin²(x) = 10 + 5cos(x) in the Interval [0, 2π)

As an excellent high school mathematics teacher, I am excited to guide you through the process of finding all solutions for the equation 10sin²(x) = 10 + 5cos(x) in the interval [0, 2π).

To begin, let's rearrange the equation to isolate the trigonometric functions:

10sin²(x) - 5cos(x) - 10 = 0

Now, we can use the trigonometric identity sin²(x) = (1 - cos(2x))/2 to rewrite the equation:

5(1 - cos(2x)) - 5cos(x) - 10 = 0 5 - 5cos(2x) - 5cos(x) - 10 = 0 5 - 5(cos(2x) + cos(x)) - 10 = 0 5 - 5(2cos(x + x) / 2) - 10 = 0 5 - 10cos(3x/2) - 10 = 0 -10cos(3x/2) = 5 cos(3x/2) = -1/2

Now, we can find the values of x that satisfy this equation. The inverse cosine function gives us:

3x/2 = π, 3π x = 2π/3, 4π/3

To find all solutions in the interval [0, 2π), we need to consider the periodicity of the trigonometric functions. Since the period of the cosine function is 2π, we can add or subtract multiples of 2π to the solutions we have found.

The solutions in the interval [0, 2π) are:

x = 2π/3, 4π/3

Therefore, the equation 10sin²(x) = 10 + 5cos(x) has two solutions in the interval [0, 2π).

I hope this detailed explanation has helped you understand the process of finding all solutions for this trigonometric equation. If you have any further questions, feel free to ask!