Question

Please find the general solutions to 6cos^2(2x)-3=0

Answer 1

\[2\cos ^2x=\frac{ 3 }{ 6 }*2=1,1+\cos 4x=1,\cos 4x=0\]
can you find the angle where cos4x=0

Answer 2

correction\[2\cos ^2(2x)\]

Answer 3

it is on y axis.

Answer 4

cos^2(2x)=1/2

Answer 5

\[\cos 4x=\cos \frac{ \left( 2n+1 \right)\pi }{ 2 },4x=\left( 2n+1 \right)\frac{ \pi }{ 2 },x=?\]
here n is an integer.

Answer 6

Hello

Answer 7

yes

Answer 8

are you able to find the solution?

Answer 9

no i made it to cos^2(2x)=1/2

Answer 10

do you square the 1/2 to get (√2/2)

Answer 11

then to 2x=pi/4+2npi

Answer 12

\[x=\frac{( 2n+1)\pi }{ 8 }\]

Answer 13

\[\cos 2x=\pm \frac{ 1 }{ \sqrt{2} }\]
to avoid negative sign i have used double angle formula.

Answer 14

ok

Answer 15

you got it.

Answer 16

got it thanks!

Answer 17

yw.