Solve for all values of X ? Sinx= cosx+1

1 Answer
VNVDVI
Mar 22, 2018

#x=pi/2+npi# where #n# is any integer.

Explanation:

We first want to get this equation in terms of only one trigonometric function. Let's say we want to put everything in terms of cosine, as that will work best in this case.

Let's square both sides:

#sin^2x=(cosx+1)^2#

#sin^2x=cos^2x+2cosx+1#

Recall the identity

#sin^2x+cos^2x=1#

Solving for #sin^2x,# we get

#sin^2x=1-cos^2x#

Thus,

#1-cos^2x=cos^2x+2cosx+1#

And everything is in terms of cosine. Let's move everything to one side:

#cos^2x+3cosx=0# (Our #1's# cancel out)

We can factor out an instance of #cosx:#

#cosx(cosx+3)=0#

We now solve the following equations:

#cosx=0#

#cosx+3=0#

For #cosx=0, x=pi/2+npi# where #n# is an integer because #cosx=0# for #pi/2, 3pi/2, 5pi/2, 7pi/2,...#, meaning values of #x# which cause #cosx=0# are infinitely many and repeat every #pi# units (so we add #npi#).

#For cosx+3=0#, let's solve for #cosx:#

#cosx=-3#

We already see no values of #x# solve this. #-1<=cosx<=1, # always. #cosx# cannot be #-3.#

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