Question

Find all the values of $\alpha$ for which the equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is valid. Also, find the general solution of the equation.

Answer 1

Here we need to substitute the appropriate algebraic identities and trigonometric identities to find the desired value of $\alpha$. We need to find the value of $\alpha$such that the given equation is valid or solvable. To find that we just need to convert the equation in either sine form or cosine form by the help of trigonometric formulas. Then we will get a quadratic equation. By solving this we will get the required answer.

Formula to be used:
a) ${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$
b) ${\sin ^2}x + {\cos ^2}x = 1$
c) $\sin x\cos x = \dfrac{{\sin 2x}}{2}$
d) The quadratic formula of the form$a{x^2} + bx + c = 0$ is as follows.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
e)If $\sin x = \sin \theta$ , then the general solution is $x = n\pi + {\left( { - 1} \right)^n}\theta$ where $\theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$

Complete step-by-step answer:
Here the given equation is ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$and we are asked to calculate the values of $\alpha$so that the given equation is solvable. Also, we need to find the general solution of the given equation.
${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$
We shall rewrite the given equation for our convenience.
$\Rightarrow {\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^2} + \sin 2x + \alpha = 0$
Now we shall apply the algebraic identity ${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$
$\Rightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x + \sin 2x + \alpha = 0$
Now we shall apply ${\sin ^2}x + {\cos ^2}x = 1$ in the above equation.
$\Rightarrow {\left( 1 \right)^2} - 2{\sin ^2}x{\cos ^2}x + \sin 2x + \alpha = 0$
$\Rightarrow 1 - 2{\left( {\sin x\cos x} \right)^2} + \sin 2x + \alpha = 0$
We know that $\sin 2x = 2\sin x\cos x$ and it can also be written as $\sin x\cos x = \dfrac{{\sin 2x}}{2}$
We need to apply $\sin x\cos x = \dfrac{{\sin 2x}}{2}$in the above equation.
$\Rightarrow 1 - 2{\left( {\dfrac{{\sin 2x}}{2}} \right)^2} + \sin 2x + \alpha = 0$
$\Rightarrow 1 - 2\dfrac{{{{\sin }^2}2x}}{4} + \sin 2x + \alpha = 0$
$\Rightarrow 1 - \dfrac{{{{\sin }^2}2x}}{2} + \sin 2x + \alpha = 0$
$\Rightarrow \dfrac{{2 - {{\sin }^2}2x + 2\sin 2x + 2\alpha }}{2} = 0$
$\Rightarrow 2 - {\sin ^2}2x + 2\sin 2x + 2\alpha = 0$ ………$\left( 1 \right)$
We know that sine lies between $- 1$ and $1$ .
Let $y = \sin 2x$
Since $- 1 \leqslant \sin 2x \leqslant 1$we have $- 1 \leqslant y \leqslant 1$
Now, we shall $y = \sin 2x$in the equation $\left( 1 \right)$
$\Rightarrow 2 - {y^2} + 2y + 2\alpha = 0$
$\Rightarrow {y^2} - 2y - \left( {2\alpha + 2} \right) = 0$
The quadratic formula of the form$a{x^2} + bx + c = 0$ is as follows.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here $a = 1$ , $b = - 2$ and $c = - \left( {2\alpha + 2} \right)$
Thus, we have $y = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 1 \times \left( { - \left( {2\alpha + 2} \right)} \right)} }}{{2 \times 1}}$
$\Rightarrow y = \dfrac{{2 \pm \sqrt {4 - 4\left( { - 2\alpha - 2} \right)} }}{2}$
$\Rightarrow y = \dfrac{{2 \pm \sqrt {4 + 8\alpha + 8} }}{2}$
$\Rightarrow y = \dfrac{{2 \pm \sqrt {4\left( {1 + 2\alpha + 2} \right)} }}{2}$
$\Rightarrow y = \dfrac{{2\left( {1 \pm \sqrt {3 + 2\alpha } } \right)}}{2}$
$\Rightarrow y = 1 \pm \sqrt {3 + 2\alpha }$
We know that $- 1 \leqslant y \leqslant 1$
Thus, we have $- 1 \leqslant 1 \pm \sqrt {3 + 2\alpha } \leqslant 1$
$\Rightarrow - 1 - 1 \leqslant \pm \sqrt {3 + 2\alpha } \leqslant 1 - 1$
$\Rightarrow - 2 \leqslant \pm \sqrt {3 + 2\alpha } \leqslant 0$
Since the interval ends at zero, we shall ignore the positive root.
$\Rightarrow - 2 \leqslant - \sqrt {3 + 2\alpha } \leqslant 0$
$\Rightarrow 2 \geqslant \sqrt {3 + 2\alpha } \geqslant 0$
$\Rightarrow 0 \leqslant \sqrt {3 + 2\alpha } \leqslant 2$
$\Rightarrow 0 \leqslant 3 + 2\alpha \leqslant {2^2}$
$\Rightarrow 0 \leqslant 3 + 2\alpha \leqslant 4$
$\Rightarrow 0 - 3 \leqslant 2\alpha \leqslant 4 - 3$
$\Rightarrow - 3 \leqslant 2\alpha \leqslant 1$
$\Rightarrow \dfrac{{ - 3}}{2} \leqslant \alpha \leqslant \dfrac{1}{2}$
Thus, $\alpha \in \left[ { - \dfrac{3}{2},\dfrac{1}{2}} \right]$
Hence, ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$is valid when $\alpha \in \left[ { - \dfrac{3}{2},\dfrac{1}{2}} \right]$
Now, we shall find the general solution of ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$
We have assumed $\sin 2x = y$ and we found $y = 1 \pm \sqrt {3 + 2\alpha }$
So, $\sin 2x = 1 \pm \sqrt {3 + 2\alpha }$
$\Rightarrow \sin 2x = \sin \left( {{{\sin }^{ - 1}}\left( {1 \pm \sqrt {3 + 2\alpha } } \right)} \right)$
If $\sin x = \sin \theta$ , then the general solution is $x = n\pi + {\left( { - 1} \right)^n}\theta$ where $\theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
$\Rightarrow \theta = {\sin ^{ - 1}}\left( {1 \pm \sqrt {3 + 2\alpha } } \right)$
Hence, the required general solution is $2x = n\pi + {\left( { - 1} \right)^n}\left( {{{\sin }^{ - 1}}\left( {1 \pm \sqrt {3 + 2\alpha } } \right)} \right)$ where $\alpha \in \left[ { - \dfrac{3}{2},\dfrac{1}{2}} \right]$
$\Rightarrow x = \dfrac{{n\pi }}{2} + \dfrac{{{{\left( { - 1} \right)}^n}{{\sin }^{ - 1}}\left( {1 \pm \sqrt {3 + 2\alpha } } \right)}}{2}$ where $\alpha \in \left[ { - \dfrac{3}{2},\dfrac{1}{2}} \right]$

Note: We are asked to find the value of $\alpha$such that the given equation is solvable. It means that we need to find the range of $\alpha$. Also, it is a well-known fact that the sine function is an increasing function and it always lies between $- 1$ and $1$ . To find the general solution, we need to know all the general solutions for all the trigonometric functions.