Question

For what values of the parameter k does the equation sin^4 x cos^4 x = k^2 have a solution?

Answer 1

Answer:

sin

4

x+cos

4

x+sin2x+α=0

⇒(sin

2

x+cos

2

x)

2

−2sin

2

xcos

2

x+sin(2x)+α=0

⇒1−2sin

2

xcos

2

x+sin(2x)+α=0⇒1−

2

sin

2

(2x)

+sin(2x)+α=0

⇒2−sin

2

(2x)+2sin(2x)+2α=0

⇒sin

2

(2x)−2sin(2x)−(2α+2)=0

Let y=sin2x, so that y=[−1,1], \sin ce −1≤sin2x≤1

Thus, y

2

−2y−(2α+2)=0

⇒y=1±

3+2a

But, −1≤y≤1, so:

⇒−1≤1±

3+2a

≤1

⇒−2≤±

3+2a

≤0

\sin ce, the range is from -2 to 0, then ignore the positive root

⇒−2≤−

3+2a

≤0

⇒0≤3+2a≤4

⇒−3≤2a≤1

⇒−

2

3

≤a≤

2

1

Therefore, Answer is −

2

3

≤a≤

2

1