Question

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

Answer 1

`143^@30 and 269^@82`

Explanation:

Rewrite the equation:
sin x - 2cos x = - 1
Call t the arc that `tan t = sin t/(cos t) = -2` --> `t = 116^@56`
`sin x - (sin t)/(cos t)cos x = -1`
`sin x.cos t - sin t.cos x = -cos t = 0.45`
`sin (x - t) = sin (x - 116.56) = 0.45`.
There are 2 solution arcs:
a. `x - 116.56 = 26^@74` --> `x = 26.74 + 116.57 = 143^@30`
b. x - 116.56 = 180 - 26.74 = 153.26 -->
`x = 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