SOLUTION: Find to the nearest tenth of a degree ,all values of x in the interval 0 less than/equal to x lessthan/equal to 360 degrees that satisfy the equation 4cos^2x-5sinx-5=0 Can you p

Algebra ->  Trigonometry-basics -> SOLUTION: Find to the nearest tenth of a degree ,all values of x in the interval 0 less than/equal to x lessthan/equal to 360 degrees that satisfy the equation 4cos^2x-5sinx-5=0 Can you p      Log On


   

Question 459274: Find to the nearest tenth of a degree ,all values of x in the interval 0 less than/equal to x lessthan/equal to 360 degrees that satisfy the equation 4cos^2x-5sinx-5=0
Can you please help me answer this?
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
First we simplify this equation:4%281-%28sinx%29%5E2%29-5sinx-5=0=>
=>4-4%28sinx%29%5E2-5sinx-5=0=>4%28sinx%29%5E2%2B5sinx%2B1=0.
In the last equation we substitute sinx=y, and get the quadratic equation:
4y%5E2%2B5y%2B1=0, solving this equation we get: y=-1 and y=-1/4.
Remembering our substitution y=+sinx we need to solve two simple
trigonometric equations equivalent with our original equation.
1)sinx=-1, where x=3%2Api%2F2, and
2)sinx=-1%2F4, where x=-0.25, and x=2%2Api%2B0.25.