solve-log-24sinx-24cosx-3-2-

Question Number 83289 by Cmr 237 last updated on 29/Feb/20

s o l v e

{l o g}_{(24 s i n x)} (24 c o s x) = \frac{3}{2}

Answered by TANMAY PANACEA last updated on 29/Feb/20

\frac{l n 24 c o s x}{l n 24 s i n x} = \frac{3}{2}

\frac{l n 24 c o s x}{3} = \frac{l n 24 s i n x}{2} = k

24 c o s x = e^{3 k}

24 s i n x = e^{2 k}

24^{2} ({c o s}^{2} x + {s i n}^{2} x) = e^{6 k} + e^{4 k}

{(e^{2 k})}^{3} + {(e^{2 k})}^{2} = 576 = 8^{3} + 8^{2}

e^{2 k} = 8 \to 2 k = l n 8

k = \frac{3 l n 2}{2} = \frac{3}{2} l n 2

n o w 24 s i n x = e^{2 k} = 8

s i n x = \frac{1}{3} x = {s i n}^{- 1} (\frac{1}{3})

★ r e c h e c k \dots e^{2 k} = 8 e^{k} = \sqrt{8}

e^{3 k} = 8 \sqrt{8} s o

24 c o s x = 8 \sqrt{8}

c o s x = \frac{\sqrt{8}}{3} ★

Answered by mr W last updated on 29/Feb/20

\frac{\ln (24 \cos x)}{\ln (24 \sin x)} = \frac{3}{2}

{(24 \cos x)}^{2} = {(24 \sin x)}^{3}

\cos^{2} x = 24 \sin^{3} x

24 \sin^{3} x + \sin^{2} x - 1 = 0

24 s^{3} + s^{2} - 1 = 0

(3 s - 1) (8 s^{2} + 3 s + 1) = 0

\Rightarrow s = \sin x = \frac{1}{3} \Rightarrow x = n π + {(- 1)}^{n} \sin^{- 1} \frac{1}{3}

\Rightarrow s = \sin x = \frac{- 3 \pm i \sqrt{23}}{16} \Rightarrow n o n - r e a l

Commented by Cmr 237 last updated on 29/Feb/20

t h k y o u s i r

Commented by jacoque@gmail.com last updated on 29/Feb/20

g o o d v e r y g o o d

Answered by jacoque@gmail.com last updated on 29/Feb/20

{(24 \sin x)}^{\frac{3}{2}} = 24 \cos x

\frac{{(\sin x)}^{\frac{3}{2}}}{\cos x} = \frac{24}{24^{\frac{3}{2}}}

\frac{{(\sin x)}^{\frac{3}{2}}}{\cos x} = \frac{\sqrt{6}}{12}

\frac{{(1 - \cos^{2} x)}^{\frac{1}{2} \times \frac{3}{2}}}{\cos x} = \frac{\sqrt{6}}{12}

\frac{{({(1 - \cos^{2} x)}^{\frac{1}{2} \times \frac{3}{2}})}^{4}}{\cos^{4} x} = \frac{{(\sqrt{6})}^{4}}{12^{4}}

\frac{{(1 - \cos^{2} x)}^{3}}{\cos^{4} x} = \frac{1}{576}

\frac{(1 - 2 \cos^{2} x + \cos^{4} x) (1 - \cos^{2} x)}{\cos^{4} x} = \frac{1}{576}

\frac{1 - \cos^{2} x - 2 \cos^{2} x + 2 \cos^{4} x + \cos^{4} x - \cos^{6} x}{\cos^{4} x} = \frac{1}{576}

\frac{1}{576}

576 - 1728 \cos^{2} x + 1728 \cos^{4} x - 576 \cos^{6} x -

- \cos^{4} x = 0

576 \cos^{6} x - 1727 \cos^{4} x + 1728 \cos^{2} x - 576 = 0

\cos x = 0.942809041582 x = 0.339836

\cos x = - 0.942809041582

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