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Mathematics
Bryant Clark
5 September, 05:24
2 cos x + 3 sin 2x = 0
answer in degrees
+3
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1
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Ivan Velazquez
5 September, 07:10
0
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.
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