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

2 Answers
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.