How to solve #3sin(x+10)=cos(x-20)#?

Question

2045 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-07T21:07:33

#x = 180n + tan^-1 gamma #

where #gamma = (cos20 - 3sin10)/(3cos10 - sin20 ) #

#n in ZZ #

The first thing to consider the additional formulea;

#cos(A+B) = cosAcosB - sinAsinB#

#sin(A+B) = sinAcosB + sinBcosA#

So hence #3sin(x+10) = cos(x-20) # becomes;

#3sinx cos 10 + 3cosx sin 10 = cosx cos 20 + sinx sin 20#

Hence we can rearange, yielding;

#(3cos10 - sin 20 ) sinx = (cos20 - 3sin20 ) cosx #

#=> tanx = (cos20 - 3sin 10 )/(3cos10 - sin20 #

Letting; #gamma = (cos20 - 3sin10)/(3cos10 - sin20 ) #

We know #gamma = tan(tan^-1 gamma ) #

So #tanx = tan(tan^-1 gamma ) #

We can now use the general solution of #tanx#:

# tan theta = tan alpha# , #=> theta = 180n + alpha, n in ZZ#

Hence #x = 180n + tan^-1 gamma #, #n in ZZ #