Solve this equation 15+21 sin (x) = 20 cos^2 (x) , where 0 degrees less than and equal to x and less than 360 degrees

Question

Solve the equation 15+21 sin (x) = 20 cos^2 (x), where 0 degrees is less than and equal to x and x is less than or equal to 360 degrees. Round to thw nearest hundredth of a degree.

Answer 1

Use the identity for cos^2(x)=1-sin^2(x)

 

Plugging this identity in gives the equation:

 

15+21sinx=20(1-sin^2(x))

 

Solve

 

15+21sinx=20(1-sin^2(x))

 

15+21sinx=20-20sin^2(x)

+20sin^2(x)      +20sin^2(x)  

 

20sin^2(x)+21sin(x)+15=20

                                -20  -20

20sin^2(x)+21sin(x)-5=0  (recognize this form? Hint: its ax^2+bx+c=0; its a quadratic)

 

So solving that quadratic isn't bad with a graphing calculator. Plug the equation into y one and find where the graph intersects zero (calculator syntax: 2nd, trace, zeros). Make sure to have your calculator set in radian mode, otherwise you won't be able to graph the equation. Set your x window from 0 to 2pi (which is the same thing as 0 to 360degrees). You will get two answers to the equation: 2.94 and .201. Remember, because have the calculator in radians, these answers are in radians. Because the question gave the domain in degrees, change the answers to degrees by multiplying radians(180/pi). This makes the final answers 168.45 and 11.51 degrees.