Question

Find all the values that satisfy
cos^2(2x)+2sin^2 (2x)=2

Answer 1

Refer to your trig-identities to convert cosine or sine to get all the same. Then subtract 2, and set equal to zero.

Solve.

Answer 2

should I make it
\[(\cos2x)^{2}+2(\sin2x)^2=2?\]

Answer 3

Well, what is cos(2x) or sin(2x)? Half-angle formula...

Answer 4

You want to get it in the form: \(\sf \color{blue}{ax^2+bx+c}\) but in trig form.

Answer 5

should I use
\[\cos2x=1-2\sin ^{2}x\]?

Answer 6

Sure, which ever is easier for you :-)

Answer 7

i got
\[-4\sin ^{4}x-7sinx+1=2?\]

Answer 8

I am stuck @abb0t

Answer 9

to me, I will break it like \[cos^2 (2x) +sin^2(2x)+sin^2(2x) =2\\1 + sin^2(2x) = 2\\sin^2(2x) = 1\\sin(2x)= \pm1\]
for sin(2x)=1 \(\rightarrow\) 2x= \(\dfrac{\pi}{2}\rightarrow x =\dfrac{\pi}{4}\) generalize it by + 2k\(\pi\)
do the same with sin(2x)= -1

Answer 10

I get it thanks