How do you solve cos(2x)cos(x)+sin(2x)sin(x)=1 for all solutions?

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Feb 26, 2018

See below...

Explanation:

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Feb 26, 2018

Apply appropriate trig identities and simplify, resulting in having to solve the equation #cos(x)=1#. This means that #x=2pin# where n is any integer.

Explanation:

First, we want everything in this equation to be in the form of one trigonometric function. Let's have everything in the form of #cos(x)#.

Recall the following identity:

#sin(2x)=2sin(x)cos(x)#

Rewrite with this applied:

#cos(2x)cos(x)+2sin(x)cos(x)sin(x)=1#

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

Recall that

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

Solving for #sin^2(x)# gives

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

Apply this to the instance of #sin^2(x)# in the equation:

#cos(2x)cos(x)+2cos(x)(1-cos^2(x))=1#

Distribute #2cos(x)# through:

#cos(2x)cos(x)+2cos(x)-2cos^3(x)=1#

Recall the following identity:

#cos(2x)=2cos^2(x)-1# (There are three ways to write the double-angle identity for #cos(2x)#; however, we only want the one where everything is in terms of #cos(x)#)

Apply:

#(2cos^2(x)-1)cos(x)+2cos(x)-2cos^3(x)=1#

Multiply #cos(x)# through and simplify:

#cancel(2cos^3(x))-cos(x)+2cos(x)-cancel(2cos^3(x))=1#

Now, we have a very simple equation to solve:

#cos(x)=1#

We want all values of #x# for which #cos(x)=1#.

#cos(x)=1# for #x=2pin# where n is any integer. We must include #n# because the period of #cos(x)# is #2pi#, meaning that any value of #cos(x)# repeats every #2pi# units (regardless of whether we add or subtract #2pi#). #cos(0)=1#, but so does #cos(2pi), cos(4pi), cos(-2pi)#, and infinitely many other values obtained; therefore, we account for these with #2pin.#