2cos x +3 sin 2x = 0
answer in degrees

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.

the first one

step-by-step explanation:

just id the test.