Can someone help me with this question. The mark scheme says its C but i have no idea why or how you get to that.
5 Answers
$\begingroup$HINT
Let $t=2^{\sin x}$ then
$$2^{2\sin x}+2^{\sin x}-6=0 \iff t^2+t-6=(t+3)(t-2)=0$$
$\endgroup$ 2 $\begingroup$As $2^{2\sin x}=(2^{\sin x})^2.$
If $2^{\sin x}=y,$ as $-1\le\sin x\le1\implies2^{-1}\le y\le2^1$
we have $$y^2+y-6=0$$
$\endgroup$ $\begingroup$This is a quadratic in the variable $2^{\sin x}$. Make the substitution $u = 2^{\sin x}$ to obtain$$ u^2 + u - 6 = 0 $$and factor, revealing$$ (u+3)(u-2) = 0 \text{.} $$This says either $2^{\sin x} = -3$ or $2^{\sin x} = 2$. Since powers of $2$ are never negative, we discard the former and continue studying the latter. For $2^{\sin x}$ to be $2$, $\sin x = 1$. (The function $2^v$ is one-to-one, meaning there is only one power of $2$ that is equal to $2$, in particular, the first power.)
Now we just have trigonometry. From$$ \sin x = 1 $$we (generically) have the two infinite families of solutions$$ x = \arcsin(1) + 2 \pi k $$and$$ x = \pi - \arcsin(1) + 2 \pi k \text{,} $$for any integer $k$. You should know $\arcsin(1) = \pi/2$, so both of these familes are actually the same family (because $\pi/2 = \pi - \pi/2$).
$\endgroup$ 1 $\begingroup$How it is stated above it the set such that $\sin (x)=1 $ although there are also non-real solutions for $2^{\sin x}=-3$. Take a look at Wolfram graphs.
$\endgroup$ $\begingroup$As a final step after the other hints:
Once you get the roots for $t=2^{\sin x}$, check which ones correspond to real values of $x$. You have to have $-1\le\sin x\le+1$ so $(1/2)\le t\le 2$. For all roots for $t$ that satisfy this result, find $\sin x$ and match that to the choices given to find the correct one. Remember that the sine is periodic and you have to identify the right period.
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