the eq:sin^4x - 2cos^2x +k^2=0 will have real solution if?

Question

A) k^2<=2
B) k^2>=1
C) k^2<=3
D) none
chose correct option and explain

Answer 1

The top solution is wrong, a crucial element is missing. Read on...

First part is right, replace cos^2 with 1-sin^2 in the equation to get:

sin^4(x) -2(1-sin^2(x))+k^2 = 0

sin^4(x) + 2sin^2(x) + (k^2-2) = 0

This is a quadratic equation in terms of sin^2(x). So using -(b±sqrt(b^2-4ac))/2a, we have

sin^2(x) = (-2 ± sqrt(4-4(k^2-2))) / 2

= -1 ± sqrt(3-k^2).

Now, sin^2(x) must be positive (it's a square) and no greater than 1 (it's a sine). So,

0 <= -1 + sqrt(3-k^2) <=1 (only the positive root will make this positive)

1 <= sqrt(3-k^2) <= 2

1 <= 3-k^2 <= 4

-2 <= -k^2 <= 1

-1 <= k^2 <= 2

The left-hand inequality is always satisfied, so the only remaining condition is k^2<=2.

Contrary to what other said, k^2<=3 is NOT a sufficient condition because it results in a negative solution for sin^2(x).

Answer 2

Let us make it in terms of sin x
sin^4 x -2(1-sin^2x) + k^2 =0
or

sin^4 x + 2 sin ^2 x -2 + k^2 = 0

or (sin^2 x + 1)^2 - 3 + k^2 = 0
k^2 = 3 -(1+ sin^2 x)^2

k^2 <= 2 as (1+ sin^2 x)^2 >= 1
the RHS minimum can be -1 but LHS minumum 0

so A is correct

Answer 3

sin^4 x - 2cos^2 x + k^2 =0 (1)
<=> (1-cos2x)^2 - 4(1+cos2x) + 4k^2=0
<=> cos^2 2x - 2cos2x - 3 + 4k^2 =0 (*)
let t = cos2x ( |t|<=1)
(*) becomes t^2 -2t - 3 +4k^2 = 0 (2)
(1) will have root x if (2) has root t sothat |t| <= 1

after solving this problem we will have k^2 <=1/4

Answer 4

c
cos^2x=1-sin^2x so you have sin^4x+2sin^2x+(k^2-2)=0
replace sin^2x=z then you have z^2+2z + (k^2-2)=0
the eq now has solution if 12-4^2>=0 which means k^2<=3