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The number of solutions of the equation $${(\sqrt3\sin x+\cos x)}^{\sqrt{\sqrt3\sin{2x}-\cos{2x}+2}}=4$$ is$\ldots$

I know that i have to sketch the graph of the left-hand side and then look at the point of intersections with the line $y=4$. But i could not got ahead with sketching the graph manually. I tried it in desmos and it shows an infinite number of solutions.

Rócherz
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Navin
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2 Answers2

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$$\sqrt3\sin2x-\cos2x+2=2\left\{1-\cos\left(2x+\frac\pi3\right)\right\}=4\sin^2\left(x+\dfrac\pi6\right)$$

$$(\sqrt3\sin x+\cos x)^{\sqrt{\sqrt3\sin2x-\cos2x+2}}=\left(2\sin\left(x+\frac\pi6\right)\right)^{\left|2\sin\left(x+\frac\pi6\right)\right|}=\left|2\sin\left(x+\frac\pi6\right)\right|^{\left|2\sin\left(x+\frac\pi6\right)\right|}$$

As $0\le\left|\sin\left(x+\frac\pi6\right)\right|\le1,$

the equality occurs iff $\left|\sin\left(x+\frac\pi6\right)\right|=1$

if $\cos\left(x+\frac\pi6\right)=0, x+\frac\pi6=(2m+1)\frac\pi2$ where $m$ is any integer

lab bhattacharjee
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The base of the exponentiation is a number between $-2$ and $2$, while the exponent is between $0$ and $2$.

So the only possible solutions, if any, are with $(-2)^2$ and $2^2$.

By solving simple trigonometric equations, this indeed occurs for $x=\dfrac{\pi}3+k\pi$.

enter image description here

The strange looking plot is due to the exponentiation of a negative, undefined for an irrational exponent and taking both signs for rational ones.