Question #f0862

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Answer 1 · 2018-02-23T14:38:29

See the explanation below

You must write

#(5sinx+1cosx)# as #(rsin(x+alpha))#

Let

#asinx+bcosx=rsin(x+alpha)#

#=r(sinxcosalpha+cosxsinalpha)#

So,

#a=rcosalpha# and

#b=rsinalpha#

#tanalpha=b/a#

#alpha=arctan(b/a)#

#a^2/r^2+b^2/r^2=1#

#r^2=a^2+b^2#

#r=sqrt(a^2+b^2)#

Therefore,

#asinx+bcosx=sqrt(a^2+b^2)sin(x+alpha)#

Applyng this

#5sinx+1cosx=sqrt26sin(x+alpha)#

Where,

#alpha=arctan(1/5)#

Then, you can solve easily your equation.