SOLUTION: Solve the trigonometric inequality. Write the solution in interval notation. 2cosē(3x) + 5cos(3x) - 3

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the trigonometric inequality. Write the solution in interval notation. 2cosē(3x) + 5cos(3x) - 3      Log On


   

Question 901911: Solve the trigonometric inequality. Write the solution in interval notation.
2cosē(3x) + 5cos(3x) - 3
Answer by Edwin McCravy(20081) About Me  (Show Source):
You can put this solution on YOUR website!
2cosē(3x) + 5cos(3x) - 3 < 0

The left factors just as 2uē+5u-3 = (u+3)(2u-1)

[cos(3x)+3][2cos(3x)-1] < 0

We find the critical values by finding the zeros of the left side

The first factor has no zeros since no cosine can be -3
The second factor has zeros when cos(3x) = 1%2F2

This is when 3x=pi%2F3+%2B2n%2Api   and when 3x=5pi%2F3%2B2n%2Api
             x=pi%2F9+%2B2n%2Api%2F3          x=5pi%2F9%2B2n%2Api%2F3 
             x=pi%2F9+%2B6n%2Api%2F9          x=5pi%2F9%2B6n%2Api%2F9
             x=expr%28pi%2F9%29%281%2B6n%29            x=expr%28%285pi%29%2F9%29%285%2B6n%29

graph%28400%2C400%2C-2pi%2C2pi%2C-7%2C5%2C2cos%283x%29%5E2%2B5cos%283x%29-3%29

From the graph of y = 2cosē(3x) + 5cos(3x) - 3 we see that the graph is negative
(below the x-axis) in all open intervals
, when n is any integer.       

Edwin