How to solve cos2x<3sinx + 2 for (-π<x<π)?
2 Answers
cos 2x < 3sin x +2
Replace cos 2x by
First, solve f(x) = 0.
Since a - b + c = 0, use shortcut. The 2 real roots are:
sin x = - 1 and
When sin x = - 1 -->
When
Within the interval
Within interval (0, pi), --> f(x) always positive because sin x > 0.
Answer: f(x) > 0 within the 3 intervals:
We can check the answer by using graphing calculator to graph the function f(x). The parts of the graph that stay above the x-axis give the answers for f(x) > 0.
Check by calculator:
Given:
Substitute
Add
Please look at
I have marked the points
We can show this analytically by factoring the quadratic:
This gives us the roots:
The two points that I have marked,
The second root is a false solution, because it does not cross the
Therefore, the solution to the inequality is these two regions: