How do you solve #sin(60-x)=2sinx#?

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Answer 1 · 2015-05-03T03:29:59

Use the trig identity: #sin (a - b) = sin a.cos b - sin b.cos a#

#sin (60 - x) = sin 60.cos x - cos 60.sin x = (sqr3/2).cos x - (sin x)/2#

(sqr3).cos x - sin x = 4.sin x#

#f(x) = (sqr3).cos x - 5sin x = 0.# Divide by cos x -->

#(Condition: cos x not 0 -> x not pi/2 and x not 3pi/2)#

#f(x) = sqr3 - 5tan x -> tan x = (sqr3)/5 = 0.35#

Calculator gives #x = 19.10 + k.pi#