Number of solution of equation $16 \sin^3x=14+(\sin x+7)^{\frac{1}{3}}$ in $[0,4\pi]$.

Question

Number of solutions of equation $16 \sin^3x=14+(\sin x+7)^{\dfrac{1}{3}}$ in $[0,4\pi]$.

My thinking:

1. Directly satisfy $\sin x=1$.

2. I thought of doing $f(x)=f^{-1}(x)\;$ but couldn't proceed further because I can't think of a function on the left side whose inverse is the right side.

Answer 1

Let $y=\sin x$. Then $$16(y^3-1)=\sqrt[3]{y+7}-2$$ $$16(y^3-1)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=y-1$$ Then $y=1$ or $$16(y^2+y+1)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=1$$ $$4(4y^2+4y+4)\left(\left(\sqrt[3]{y+7}\right)^2+2\sqrt[3]{y+7}+4\right)=1$$ $$4\left((2y+1)^2+3\right)\left(\left(\sqrt[3]{y+7}+1\right)^2+3\right)=1$$ But $$4\left((2y+1)^2+3\right)\left(\left(\sqrt[3]{y+7}+1\right)^2+3\right) \ge 4 \cdot 3 \cdot 3= 36 >1$$. Only solution $y=1$

Answer 2

Let $s = \sin x$. Then your equation is $16s^3 = 14 + (s + 7)^{1/3}$. Equivalently,

$$4096s^9 - 10752s^6 + 9408s^3 - s - 2751 = 0$$

By the Fundamental Theorem of Algebra, there could be up to 9 possible values of $s$, but I'm assuming that you're only interested in real solutions, of which $s = 1$ turns out to be the only one.

So the question then becomes “How many numbers in $[0, 4\pi]$ have a sine of 1?”. And there are two of those: $x = \frac{\pi}{2}$ and $x = \frac{5\pi}{2}$.