Question

Solve the expression: $\cos 2x + 7 = a(2 - \sin x)$ can have a real solution for
A.All values of $a$
B.$a \in \left[ {2,6} \right]$
C.$a \in \left[ { - \infty ,2} \right]$
D.$a \in \left[ {0,\infty } \right]$

Answer 1

Hint: In this question we have been given $\cos 2x + 7 = a(2 - \sin x)$ . We can see that we have trigonometric functions in this expression, so we will try to apply the trigonometric formula to simplify this.
We will use the formula
$\cos 2\theta = 1 - 2{\sin ^2}\theta $ , by comparing here we have
$\theta = x$ , so we can write
$\cos 2x = 1 - 2{\sin ^2}x$ .

Complete step-by-step answer:
We have been given
$\cos 2x + 7 = a(2 - \sin x)$ .
We will put this value
$\cos 2x = 1 - 2{\sin ^2}x$ in the above expression, so we have:
$1 - 2{\sin ^2}x + 7 = 2a - a\sin x$
We can transfer all the terms to the left hand side, we have:
$1 - 2{\sin ^2}x + 7 - 2a + a\sin x = 0$
By arranging the terms we have:
$ - 2{\sin ^2}x + a\sin x + 7 + 1 - 2a + = 0$
We will change the sign without changing its original form, so we have:
$2{\sin ^2}x - a\sin x + 2a - 8 = 0$
Now we can see that the above expression is in the form of a quadratic form:
$a{x^2} + bx + c = 0$
We will apply the quadratic formula i.e.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , we will now put the values in the formula and we have:
$a = 2,b = - a,c = 2a - 8$
By substituting the values, we have;
$\sin x = \dfrac{{ - ( - a) \pm \sqrt {{a^2} - 4 \times 2 \times (a - 8)} }}{{2 \times 2}}$
Simplifying the values we have:
$\sin x = \dfrac{{a \pm \sqrt {{a^2} - 8 \times 2(a - 4)} }}{4}$
We have taken the common factor out, and now by multiplying we have:
$\sin x = \dfrac{{a \pm \sqrt {{a^2} - 16a - 64} }}{4}$
On simplifying and removing the square root we have:
$\sin x = \dfrac{{a \pm (a - 8)}}{4}$
We will now solve this, we can write
$\sin x = \dfrac{{a + (a - 8)}}{4}$
It gives us
$\dfrac{{2a - 8}}{4} = \dfrac{{2(a - 4)}}{4}$
So it gives the value
$\sin x = \dfrac{{a - 4}}{2}$
Another value, we can take the negative sign i.e.
$\sin x = \dfrac{{a - (a - 8)}}{4}$
On simplifying, it gives:$\dfrac{8}{4} = 2$
We know that the value $\sin x$ cannot be $2$ , because the value of $\sin x$can never exceed $1$ .
So the correct value is
$\sin x = \dfrac{{a - 4}}{2}$
Now we know that the value of sine lies between $ - 1$ to $1$ .
It can be written as
$ - 1 \leqslant \sin x \leqslant 1$
By putting the value we have:
$ - 1 \leqslant \dfrac{{a - 4}}{2} \leqslant 1$
By multiplying both the sides by $2$ , we have
$ - 2 \leqslant a - 4 \leqslant 2$
We will now add $4$ on both the sides of equation i.e.
$4 - 2 \leqslant a - 4 + 4 \leqslant 2 + 4$
It gives us value:
$2 \leqslant a \leqslant 6$
Therefore it gives:
$a \in \left[ {2,6} \right]$
Hence the correct option is (b) $a \in \left[ {2,6} \right]$
So, the correct answer is “Option b”.

Note: WE should note that in the above solution,
$\sin x = \dfrac{{a \pm \sqrt {{a^2} - 16a - 64} }}{4}$ , we know that
${a^2} - 16a + 64 = {a^2} - 2 \times a \times 8 + {(8)^2}$ .
So we will apply the formula i.e.
${(a - b)^2} = {a^2} - 2ab + {b^2}$
So by applying this we can write
${(a - 8)^2}$ .
By putting this we have: $\sin x = \dfrac{{a \pm \sqrt {{{(a - 8)}^2}} }}{4} \Rightarrow \dfrac{{a \pm (a - 8)}}{4}$