(A) Let $t = \sin^2 x$. Since $\sin x \in [-1, 1]$, we have $t \in [0, 1]$.The equation becomes $t^2 - (p+2)t - (p+3) = 0$.Factoring the quadratic equ

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(A) Let $t = \sin^2 x$. Since $\sin x \in [-1, 1]$, we have $t \in [0, 1]$.The equation becomes $t^2 - (p+2)t - (p+3) = 0$.Factoring the quadratic equation:$t^2 - (p+2)t - (p+3) = (t - (p+3))(t + 1) = 0$.Thus, the roots are $t = p+3$ or $t = -1$.Since $t = \sin^2 x$ must be in the interval $[0, 1]$, we discard $t = -1$.Therefore, we must have $0 \leq p+3 \leq 1$.Subtracting $3$ from all parts, we get $-3 \leq p \leq -2$.Thus, $p \in [-3, -2]$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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If the equation $\sin^4 x - (p+2) \sin^2 x - (p+3) = 0$ has a solution, then $p$ must lie in the interval:

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A
$[-3, -2]$
B
$(-3, -2)$
C
$(2, 3)$
D
$[-5, -3]$

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Trigonometrical Equations

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Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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If the equation $\sin^4 x - (p+2) \sin^2 x - (p+3) = 0$ has a solution, then $p$ must lie in the interval:

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