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

User Supertux
by
8.0k points

1 Answer

3 votes

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

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.

User Krist
by
8.1k points

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