3x^2-2+2sinx=0?

Question

Hey, so I'm kinda stuck on this simple-looking math problem. Can anyone show me the steps please? I tried looking for online help but I couldn't find any similar problems. I just need this equation solved for x. Thanks :]

Answer 1

This is a transcendental equation which can only be solved numerically, but we can do a little manipulation to get an idea where to start:
3x^2 - 2 + 2sinx = 0
2sinx = 2 - 3x^2
sinx = 1 - (3/2)x^2
- 1 ≤ sinx ≤ 1, so
- 1 ≤ 1 - (3/2)x^2 ≤ 1
2 ≥ (3/2)x^2 ≥ 0
4 ≥ 3x^2 ≥ 0
4/3 ≥ x^2 ≥ 0
1.1547 ≥ x ≥ - 1.1547
As it turns out,
x ≈ - 1.126299941

Answer 2

You have both an algebraic (x^2) and a transcendental (sin x) function in the same equation. Except in unusual circumstances, these defy analytic solutions. About the best you can do is an iterative solution. For instance, if you make x = 0, the expression on the left evaluates to -2. If you make x = 1, it evaluates to about 2.68. Since the function is continuous from 0 to 1, there must be a 0 between 0 and 1. So: try x = .5. That'll tell you whether the 0 is between 0 and .5 or between .5 and 1. Continue to narrow it down until you get the accuracy you're looking for.

There may, of course, be more than one solution, but you can be pretty sure they'll all be "close" to 0. As x grows, the x^2 will dominate the sinx pretty quickly.