Answer:
If want just the approximated solutions in the interval from 0 to 360:
199.47
340.53
90
270
If you want all the approximated solutions:
199.47+360k
340.53+360k
90+360k
270+360k
Step-by-step explanation:
2 cos(x)+3 sin(2x)=0
First step: Use double angle identity for sin(2x). That is, use, sin(2x)=2sin(x)cos(x).
2 cos(x)+3*2sin(x)cos(x)=0
2 cos(x)+ 6sin(x)cos(x)=0
Factor the 2cos(x) out, like so:
2cos(x)[ 1 + 3 sin(x)]=0
In order for this product to be zero, we must find when both factors are 0.
2cos(x)=0 or 1+3sin(x)=0
Let's do 2cos(x)=0 first.
2cos(x)=0
Divide both sides by 2:
cos(x)=0
So the x-coordinate is 0 on the unit at x=90 deg and x=270 deg (in the first rotation).
Let's do 1+3sin(x)=0.
1+3sin(x)=0
Subtract 1 on both sides:
3sin(x)=-1
Divide both sides by 3:
sin(x)=-1/3
Unfortunately this is not on the unit circle so I'm just going to take sin^-1 or arsin on both sides (this is the same thing sin^-1 or arsin).
x=arcsin(-1/3)=-19.47 degrees
So that means -(-19.47)+180 is also a solution so 19.47+180=199.47 .
And that 360+-19.47 is another so 360+-19.47=340.53 .
So the solutions for [0,360] are
199.47
340.53
90
270
If you want all the solutions just add +360*k to each line where k is an integer.
