SOLUTION: Find all solutions of the equation on the interval [0,2pi) -4sinx = -cos^2x+4 Write answer in terms of pi

Algebra ->  Trigonometry-basics -> SOLUTION: Find all solutions of the equation on the interval [0,2pi) -4sinx = -cos^2x+4 Write answer in terms of pi      Log On


   

Question 1160202: Find all solutions of the equation on the interval [0,2pi)
-4sinx = -cos^2x+4
Write answer in terms of pi
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find all solutions of the equation on the interval [0,2pi)
-4sin%28x%29+=+-cos%5E2%28x%29%2B4

cos%5E2%28x%29-4sin%28x%29+-4=+0............use the following identity : cos%5E2+%28x+%29=1-sin%5E2%28x%29
1-sin%5E2%28x%29-4sin%28x%29+-4=+0
-sin%5E2%28x%29-4sin%28x%29+-3=+0......both sides multiply by -1
sin%5E2%28x%29%2B4sin%28x%29+%2B3=+0.........let+sin%28x%29=u
u%5E2%2B4u+%2B3=+0.......factor
u%5E2%2Bu%2B3u+%2B3=+0
%28u%5E2%2Bu%29%2B%283u+%2B3%29=+0
u%28u%2B1%29%2B3%28u+%2B1%29=+0
%28u+%2B+1%29+%28u+%2B+3%29+=+0
solutions:
u=-1
u=-3
substitute u back
sin%28x%29=-1-> general solutions for sin+%28x+%29=-1: x=3pi%2F2%2B2pi%2An
for interval 0%3C=x%3C2pi will be => highlight%28x=3pi%2F2%29
sin%28x%29=-3-> sin%28x%29 can't be smaller than -1 for real solutions => no solution exist

MSP32.gif





Answer by ikleyn(52867) About Me  (Show Source):
You can put this solution on YOUR website!
.

-4sin(x) = -cos^2(x) + 4


Replace  cos^2(x)  by  1 - sin^2(x).  You will get


-4sin(x) = -(1-sin^2(x)) + 4


sin^2(x) + 4sin(x)x + 3 = 0


Factor


(sin(x) + 3)*(sin(x) + 1) = 0


The only real solution is  sin(x) = -1,  which gives  x = 3pi%2F2.    ANSWER







Plot y = -4sin(x) (red)  and  y = -cos^2(x) +4 (green)