how to solve a trignometric quadratic equation?

Question

I am stuck in a question... the questions says $sin^4 x -(k+2)sin^2 x -(k+3)=0$ has a solution then what is the interval in which k must lie. I tried to solve it by putting $sin^2 x =p$ then putting In the equation and then making the discriminant=0 then solving for $b^2 - 4ac=0$ and then got k =-4 but now I don't know what to do next. :/

Answer 1

$$\sin^4x-(k+2)\sin^2x-(k+3)=0$$

$$\implies\sin^2x=\dfrac{k+2\pm\sqrt{(k+2)^2+4(k+3)}}2=\dfrac{k+2\pm(k+4)}2=k+3,-1$$

As $\sin^2x\ge0,\sin^2x=-1$ is untenable

If $\sin^2x=k+3,0\le k+3\le1$

Answer 2

Hint:

You don't have to calculate the roots to answer. This is a typical exercice on quadratic equations with parameters:

Given the quadratic equation $u^2-(k+2)u-(k+3)=0$, find for which values of $k$ this equation has at least one root in the interval $[0,1]$.

There's a first condition: $\Delta \ge 0$.

Then, denote $\alpha\le\beta$ the roots of this equation, you have to be in one of the following configurations:

• $\alpha < 0 <\beta< 1$,
• $0<\alpha<\beta <1$,
• $0<\alpha<1<\beta$,
• one of the roots is $0$ or $1$

The first and third configurations translate as $p(0)< 0,\;p(1)> 0$ and $p(0)> 0,\;p(1)< 0$ respectively. This is equivalent to the condition $$p(0)p(1)< 0.$$

The second configuration is equivalent to $$p(0)p(1)>0\quad \text{and} \quad 0<(\alpha+\beta)/2=(k+2)/2<1.$$