The equation sin ^4 x-(k+3) sin ^2 x-k-4=0 ha | vedclass

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(B) Given the equation: $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$. This is a quadratic equation in terms of $\sin ^2 x$. Using the quadratic formula,$\sin ^2

Vedclass · Accepted Answer

$-4 \leq k \leq -3$

Vedclass · Answer

(B) Given the equation: $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$. This is a quadratic equation in terms of $\sin ^2 x$. Using the quadratic formula,$\sin ^2 x = \frac{(k+3) \pm \sqrt{(k+3)^2 + 4(k+4)}}{2}$. Simplifying the discriminant: $(k+3)^2 + 4k + 16 = k^2 + 6k + 9 + 4k + 16 = k^2 + 10k + 25 = (k+5)^2$. So,$\sin ^2 x = \frac{(k+3) \pm (k+5)}{2}$. This gives two possible values: $\sin ^2 x = \frac{2k+8}{2} = k+4$ or $\sin ^2 x = \frac{-2}{2} = -1$. Since $\sin ^2 x$ cannot be negative,we must have $\sin ^2 x = k+4$. Given that $0 \leq \sin ^2 x \leq 1$,we have $0 \leq k+4 \leq 1$. Subtracting $4$ from all parts,we get $-4 \leq k \leq -3$.

The equation $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$ has a solution if

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