How do you solve 2cos^2x+3cosx=-12cos2x+3cosx=−1?

1 Answer
ProfLayton
Apr 14, 2017

x=-30^ox=−30o or x=180^ox=180o

Explanation:

For simplicity, we let u=cosxu=cosx:
2cos^2x+3cosx=-12cos2x+3cosx=−1
2u^2+3u=-12u2+3u=−1
2u^2+3u+1=02u2+3u+1=0
And you'll notice that we can split the middle term and factorize
2u^2+2u+u+1=02u2+2u+u+1=0
2u(u+1)+(u+1)=02u(u+1)+(u+1)=0
(2u+1)(u+1)=0(2u+1)(u+1)=0
And now it becomes obvious (by a process formally known as Null Factor Theorem) that:
2u+1=02u+1=0 OR u+1=0u+1=0 (they can't be true at the same time, its one or the other)

Solving for each one:

2u+1=02u+1=0
2u=-12u=−1
u=-1/2u=−12
cosx=-1/2cosx=−12
And if you solve the above using your calculator, you get #x=-30^o

As for the other case:
u+1=0u+1=0
u=-1u=−1
cosx=-1cosx=−1
And using a calculator you get: x=180^ox=180o

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