Question

#13 @AZ

Answer 1

There's a power-reducing identity

\(\sin^2 u = \dfrac{1 - \cos(2u)}{2}\)

Answer 2

so ^^^ times 2 equals 1/2

Answer 3

1/2 = cos(2x)

Answer 4

how?

\( 2 \times \dfrac{1-\cos(2\cdot 2x)}{2} = \dfrac{1}{2}\)

Answer 5

so
1/2 = cos(4x)

Answer 6

oh wait my bad,

it should be equal to 1 not 1/2

so cos(4x) = 1

Answer 7

your question has 1 in it, if you check your screenshot

Answer 8

But there's a 2 before the sin^2(2x) so it's still 1/2

Answer 9

wait nvmd

Answer 10

...


\( 2\sin^2(2x) = 1\)

\( 2 \left( \dfrac{1-\cos(2\cdot 2x)}{2}\right) = 1\)

\( \cancel{2}\left( \dfrac{1-\cos(2\cdot 2x)}{\cancel{2}}\right) = 1\)

Answer 11

e.e

Answer 12

how'd you do the cross out thing

Answer 13

but ye
1-cos(4x)=1
cos(4x)=0

Answer 14

and pi/2 and 3pi/2 is when cos =0

Answer 15

so then pi/2 + 2pik

Answer 16

but remember, it's cos(4x)

so if we said cos(u) = 0

then u = pi/2 + \(\pi\)k

but then u = 4x

so you have to divide by 4

Answer 17

it's every pi, not 2pi

Answer 18

so
pi/8 +pi k

Answer 19

you also have to divide pi*k by 4 as well

Answer 20

kk