15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T601,

Given the equation $\sin ^4 x-(p+2) \sin ^2 x-(p+3)=0$ has a solution.

Let $y=\sin ^2 x$. Since, $0 \leq \sin ^2 x \leq 1$, we must have $0 \leq y \leq 1$.

The equation becomes a quadratic in

$$ y: y^2-(p+2) y-(p+3)=0 $$

Solve for $y$ using the quadratic formula

$$ \begin{aligned} & y=\frac{-B \pm \sqrt{B^2-4 A C}}{2 A} \\ & y=\frac{(p+2) \pm \sqrt{(p+2)^2-4(1)(-(p+3))}}{2} \\ & y=\frac{(p+2) \pm \sqrt{(p+4)^2}}{2} \\ & y=\frac{(p+2) \pm|p+4|}{2} \end{aligned} $$

Consider two cases for $|p+4|$

Case I $p+4 \geq 0 \Rightarrow p \geq-4$

The two roots are

$$ y_1=\frac{p+2+(p+4)}{2}=\frac{2 p+6}{2}=p+3 $$

$$ y_2=\frac{p+2-(p+4)}{2}=\frac{-2}{2}=-1 . $$

For a solutions to exist, at least one root must be in the interval $[0,1]$

$y_2=-1$ is not in $[0,1]$

for $y_1=p+3$ to be in $[0,1]$

$$ \begin{aligned} 0 & \leq p+3 \leq 1 \\ 0 & \leq p+3 \Rightarrow p \geq-3 \\ p+3 & \leq 1 \Rightarrow p \leq-2 . \end{aligned} $$

Combining these with $p \geq-4$, the valid range for $p$ is $[-3,-2]$

Case II $p+4<0 \Rightarrow p<-4$

The two roots are

$$ \begin{aligned} y_1 & =\frac{p+2-(p+4)}{2}=\frac{-2}{2}=-1 \\ y_2 & =\frac{p+2+(-(p+4))}{2} \\ & =\frac{p+2-p-4}{2}=\frac{-2}{2}=-1 \end{aligned} $$

Both roots are $y=-1$, which is not in $[0,1]$. Thus there are no solution for $p<-4$.

Combining the results from both cases, the only interval for $p$ that yields a valid solutions is $[-3,-2]$

Trigonometric Functions And Equations

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