How can you show that f(x) =2 (sin(x))^2-|x| has 5 solutions on interval [-pi,pi]?

Answer 1

(If allowed you could start with a graph. To get an idea of what is going on.)

Whether you use a graph or not, you should be able to recognize that this function is an even function. This can be checked by verifying that neither of the terms, that is the square of sin x nor the absolute value of x changes value when you replace x by -x. Therefore roots between 0 to pi will have matching negative roots between 0 and -pi.

Zero itself is a root.

What remains is to identify how many solutions there are between zero and pi. Obviously this function is continuous everywhere. For small angles, sin x is well approximated by x so for small positive angles this function is similar to x² - x which factors into x(1-x) which has roots at zero and 1. We would expect one of the roots to be near but not exactly one. The main point is this function is approximated by a upward facing quadratic with a root at zero and a root near one. So the function initially decreases and then has another root making it positive again. As sin^x is never greater than 1 in magnitude, its square is never greater than one either so as x increases, the -x becomes increasingly dominant. We would expect this graph to decrease as x grows. We know that sinπ is zero so by the time x reaches π the function is already equal to negative π. Therefore there must be another root between the positive region just after the root near one and π, at which the function transitions from positive back to negative.

We could also take the derivative of the function so see how many local extrema there are.

The derivative of this function (for positive x) ends up being 4 sin(x) cos(x) - 1 for x>0. which can be considered to be 2 sin (2x) - 1. The only positive values of x at which this equals zero are when sin(2x) = 1/2. The reference angle with a sin of 1/2 is 30 degrees (or π/6 radians) therefore the solutions for x are 15° (π/12 radians) (e.g. half of 30°|half of π/6 radians) or 75° (e.g. half of 180° - 30° = 150° or half of (π - π/6 = 5π/6) = 5π/12 ). The 15° (π/12 radians) is the location of a local minimum just to the right of the origin and the 75° (5π/12 radians) is the location of the local maximum.