The number of solutions of the equation sqrt[ | vedclass

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(C) Let $x = \sin 	heta$. The equation is $(x-1)^{1/3} + x^{1/3} + (x+1)^{1/3} = 0$.Rearranging,we get $(x+1)^{1/3} + (x-1)^{1/3} = -x^{1/3}$.Cubing

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let $x = \sin 	heta$. The equation is $(x-1)^{1/3} + x^{1/3} + (x+1)^{1/3} = 0$.Rearranging,we get $(x+1)^{1/3} + (x-1)^{1/3} = -x^{1/3}$.Cubing both sides: $(x+1) + (x-1) + 3(x+1)^{1/3}(x-1)^{1/3}((x+1)^{1/3} + (x-1)^{1/3}) = -x$.Since $(x+1)^{1/3} + (x-1)^{1/3} = -x^{1/3}$,we substitute:$2x + 3(x^2-1)^{1/3}(-x^{1/3}) = -x$.$3x = 3x^{1/3}(1-x^2)^{1/3}$.Cubing again: $27x^3 = 27x(1-x^2)$.$x^3 = x - x^3 \Rightarrow 2x^3 - x = 0$.$x(2x^2 - 1) = 0$.So,$\sin 	heta = 0$ or $\sin^2 	heta = 1/2$.For $\sin 	heta = 0$,$	heta \in \{0, \pi, 2\pi, 3\pi, 4\pi\}$ ($5$ solutions).For $\sin^2 	heta = 1/2$,$\sin 	heta = \pm 1/\sqrt{2}$.In $[0, 4\pi]$,$\sin 	heta = 1/\sqrt{2}$ has $4$ solutions and $\sin 	heta = -1/\sqrt{2}$ has $4$ solutions.Total solutions = $5 + 4 + 4 = 13$. However,checking the original equation,only $\sin 	heta = 0$ satisfies it. Thus,there are $5$ solutions.

The number of solutions of the equation $\sqrt[3]{\sin \theta - 1} + \sqrt[3]{\sin \theta} + \sqrt[3]{\sin \theta + 1} = 0$ in the interval $[0, 4\pi]$ is:

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