Question 797443: Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater than x:
sin(2x)+cos(3x)=0
Here is what I have so far:
2sin(x)cos(x)+ cos(x+2x)=0
2sin(x)cos(x) +cos(x)cos(2x)- sin(x)sin(2x)=0
2sin(x)cos(x) +cos(x)[2cos^2(x-1)]- sin(x)[2sin(x)cos(x)]=0
2sin(x)cos(x) +2cos^3(x)- cos(x)- 2sin^2(x)cos(x)=0
cos(x)[2sin(x) +2cos^2(x)-1-2sin^2(x)]=0
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater than x:
sin(2x)+cos(3x)=0
Here is what I have so far:
2sin(x)cos(x)+ cos(x+2x)=0
2sin(x)cos(x) +cos(x)cos(2x)- sin(x)sin(2x)=0
2sin(x)cos(x) +cos(x)[2cos^2(x-1)]- sin(x)[2sin(x)cos(x)]=0
2sin(x)cos(x) +2cos^3(x)- cos(x)- 2sin^2(x)cos(x)=0
cos(x)[2sin(x) +2cos^2(x)-1-2sin^2(x)]=0
------------
cos(x) = 0
--> x = pi/2, 3pi/2
---------
2sin(x) +2cos^2(x)- 1 - 2sin^2(x) = 0
2sin(x) +2(1-sin^2(x))- 1 - 2sin^2(x) = 0
2sin(x) +2 - 2sin^2(x))- 1 - 2sin^2(x) = 0
2sin(x) + 1 - 4sin^2(x)) = 0
4sin^2 - 2sin - 1 = 0
sub for sin(x), solve the quadratic, etc.
|
|
|
