Question

1) Find the value of x [-pi,pi]
\[6\sin^2(2x)+5\cos[2x]=2\]
2) Find the value of x [0,2pi]
1+cosx=sinx

Answer 1

Where are you stuck?

Answer 2

For both, I'm not sure how to start them. Any hints? @Loser66

Answer 3

replace sin^2 =1-cos^2 , and solve quadratic with variable is cos.

Answer 4

for 2) square both sides, simplify a little bit and solve for cos.

Answer 5

For the second question, I got x=pi/2 and 3pi/2??
I didn't quite get the first question... @Loser66

Answer 6

https://www.desmos.com/calculator/yn2gyvrbfe
They clearly intersect at pi/2 and pi.
When I worked out the problem I got solutions of pi/2, pi and 3pi/2.

3pi/2 is not a solution and I'm trying to understand why.
I think because we did some squaring business, it may have messed things up. We have to check for extraneous solutions perhaps?
Plugging 3pi/2 back into the function doesn't hold up.

Answer 7

Still need some assistance on the first one? :o

Answer 8

Yes, please. I need some assistance for the first question. @zepdrix

Answer 9

\[\Large\rm 6\color{orangered}{\sin^2(2x)}+5\cos[2x]=2\]Applying our Square Identity to the Orange gives us,\[\Large\rm 6\color{orangered}{(1-\cos^2(2x))}+5\cos[2x]=2\]Replace cos(2x) with some variable so it's easier to work with. Say \(\Large\rm u=\cos(2x)\)
So our problem is actually:\[\Large\rm 6(1-u^2)+5u=2\]You have a quadratic involving u.
Solve for u by some method, factoring, quadratic formula, whatever you gotta do.
You'll get some solutions, then undo the u thing.
Set cos(2x) equal to those solutions, and figure out your angle from there.