How do you solve (sinx+1)-2cosx=0(sinx+1)−2cosx=0?

1 Answer
Nghi N.
Jun 22, 2016

143^@30 and 269^@82143∘30and269∘82

Explanation:

Rewrite the equation:
sin x - 2cos x = - 1
Call t the arc that tan t = sin t/(cos t) = -2tant=sintcost=−2 --> t = 116^@56t=116∘56
sin x - (sin t)/(cos t)cos x = -1sinx−sintcostcosx=−1
sin x.cos t - sin t.cos x = -cos t = 0.45sinx.cost−sint.cosx=−cost=0.45
sin (x - t) = sin (x - 116.56) = 0.45sin(x−t)=sin(x−116.56)=0.45.
There are 2 solution arcs:
a. x - 116.56 = 26^@74x−116.56=26∘74 --> x = 26.74 + 116.57 = 143^@30x=26.74+116.57=143∘30
b. x - 116.56 = 180 - 26.74 = 153.26 -->
x = 153.26 + 116.56 = 269^@82x=153.26+116.56=269∘82

Check by calculator:
x = 143.30 --> sin x = 0.60 --> 2cos x = 1.60
0.60 - 1.60 = -1. OK
x = 269.82 --> sin x = -1 --> 2cos x = 0
-1 - 0 = -1 . OK

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