How do you solve ${cos}^{2} x = \sqrt[3]{cos x}$ $cos^2x = root(3)(cosx)$ on $[0, 2 π)$ $[0, 2pi)$ ?

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Answer 1 · 2016-11-05T01:28:41

${cos}^{2} x = \sqrt[3]{cos x}$ $cos^2x = root(3)(cosx)$

${({cos}^{2} x)}^{3} = {(\sqrt[3]{cos x})}^{3}$ $(cos^2x)^3 = (root(3)(cosx))^3$

${cos}^{6} x = cos x$ $cos^6x = cosx$

${cos}^{6} x - cos x = 0$ $cos^6x - cosx= 0$

${cos}^{5} x (cos x - 1) = 0$ $cos^5x(cosx- 1 ) = 0$

${cos}^{5} x = 0 and cos x = 1$ $cos^5x = 0 and cosx = 1$

$x = \frac{π}{2}, \frac{3 π}{2}, 0$ $x = pi/2, (3pi)/2, 0$

Hopefully this helps!

How do you solve cos^2x = root(3)(cosx)cos2x=3√cosxcos^2x = root(3)(cosx) on [0, 2pi)[0,2π)[0, 2pi)?