Question

Find all solutions to the equation
sin^2(x)cos^2(x) = 1/4.

Answer 1

cos(2x) - sin(x) = 1
cos²(x) - sin²(x) - sin(x) = 1
1 - cos²(x) + sin²(x) + sin(x) = 0
2sin²(x) + sin(x) = 0
sin(x)(2sin(x) + 1) = 0

sin(x) = 0
x = nπ

OR

sin(x) = -1/2
x = (4n-1)π/2 ± π/3

Between 0 and 2π:
0, π, 7π/6, 11π/6, 2π

Answer 2

sin^2(x)cos^2(x) = 1/4
(1-cos^2(x))cos^2(x) = 1/4
-cos^4(x)+cos^2(x)-1/4=0
cos^(2)x=1/2
cosx=sqrt(2)/2

x=45,315

Answer 3

hey @.Sam. i am learning with the asker here and i don't get


-cos^4(x)+cos^2(x)-1/4=0
cos^(2)x=1/2
cosx=sqrt(2)/2

pls explain for me? :)

Answer 4

sin^2(x)cos^2(x) = 1/4.
(1/4)(2sin cos)^2 =1/4
2sin cos = 1 or -1
sin2x =1 or -1
Could it be done like this?

Answer 5

ya it can be its near 2 the ans

Answer 6

-cos^4(x)+cos^2(x)-1/4=0
Factor,
let u=cos x
-u^(2)+u-(1)/(4)=0

u^(2)-u+(1)/(4)=0

u=(-b+-sqrt(b^(2)-4ac))/(2a) where au^(2)+bu+c=0

u=(-(-1)+-sqrt((-1)^(2)-4(1)((1)/(4))))/(2(1))

u=(1+-sqrt((-1)^(2)-4(1)((1)/(4))))/(2(1))

u=1/2

cos^(2)x=1/2

Answer 7

hey @Coco..how did you have 1/4 in the left hand side? your transitions are fast :p sorry...just want to learn along with the asker here hehe..i think he doesnt get either

Answer 8

oh ok! sam :D thanks

Answer 9

sin 2x = 2sinx cos x
we have sin^2(x)cos^2(x) =(sinx cos x)^2
in order to express it in sin2x, (2 sinx cosx /2)^2 =(1/4) (sin2x)^2