Given-sin-x-1-3-cos-x-1-3-1-3-1-3-then-sin-2-x-

Question Number 128681 by bemath last updated on 09/Jan/21

Given \sqrt[3]{\sin x} - \sqrt[3]{\cos x} = \frac{1}{\sqrt[3]{3}}

then \sin^{2} x = ?

Commented by MJS_new last updated on 09/Jan/21

not funny to solve exactly but possible \dots

Answered by MJS_new last updated on 09/Jan/21

\sqrt[3]{a} - \sqrt[3]{b} = \sqrt[3]{c} ∣^{3}

a - b - 3 \sqrt[3]{a b} (\sqrt[3]{a} - \sqrt[3]{b}) = c

[insert from above \sqrt[3]{a} - \sqrt[3]{b} = \sqrt[3]{c}]

3 \sqrt[3]{a b c} = a - b - c

27 a b c = {(a + b - c)}^{3}

a = \sin x \land b = \cos x \land c = \frac{1}{3}

9 \sin x \cos x = {(- \frac{1}{3} + \sin x - \cos x)}^{3}

let t = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan t and transforming

t^{6} + \frac{279}{4} t^{5} + 21 t^{4} - 9 t^{3} - 42 t^{2} - \frac{99}{4} t - 8 = 0

amazingly this can be completely solved

I get 3 square factors

f_{1} (t) f_{2} (t) f_{3} (t) = 0

f_{1} (t) = t^{2} + \frac{3}{4} t + \frac{1}{4}

f_{2} (t) = t^{2} + \frac{3 (23 - 3 \sqrt{57})}{2} t - \frac{67 - 9 \sqrt{57}}{2}

f_{3} (t) = t^{2} + \frac{3 (23 + 3 \sqrt{57})}{2} t - \frac{67 + 9 \sqrt{57}}{2}

only f_{3} (t) = 0 has real roots

t_{1} = - \frac{3 (23 + 3 \sqrt{57}) + \sqrt{2 (4957 + 657 \sqrt{57}}}{4} \approx - 69.4459

t_{2} = - \frac{3 (23 + 3 \sqrt{57}) - \sqrt{2 (4957 + 657 \sqrt{57}}}{4} \approx . 971609

\sin^{2} x = {(\frac{2 t}{t^{2} + 1})}^{2}

now the rest is easy

Commented by bemath last updated on 10/Jan/21

��