solve-cos-3-x-sin-2-x-7-8-adopted-from-youtube-

Question Number 164671 by mnjuly1970 last updated on 20/Jan/22

s o l v e

{c o s}^{3} (x) + {s i n}^{2} (x) = \frac{7}{8}

a d o p t e d f r o m y o u t u b e \dots

Commented by MJS_new last updated on 20/Jan/22

for 0 ⩽ x < 2 π I get

x \in {\frac{π}{5}, \frac{π}{3}, \frac{3 π}{5}, \frac{7 π}{5}, \frac{5 π}{3}, \frac{9 π}{5}}

Answered by bobhans last updated on 20/Jan/22

\cos^{3} x + 1 - \cos^{2} x - \frac{7}{8} = 0

\cos^{3} x - \cos^{2} x + \frac{1}{8} = 0

8 \cos^{3} x - 8 \cos^{2} x + 1 = 0

(2 \cos x - 1) (4 \cos^{2} x - 2 \cos x - 1) = 0

(1) \cos x = \frac{1}{2} \Rightarrow x = \pm \frac{π}{3} + 2 k π

(2) 4 \cos^{2} x - 2 \cos x - 1 = 0

\cos^{2} x - \frac{1}{2} \cos x - \frac{1}{4} = 0

{(\cos x - \frac{1}{4})}^{2} - \frac{5}{16} = 0

\cos x - \frac{1}{4} = \pm \frac{\sqrt{5}}{4}

\cos x = \frac{1 \pm \sqrt{5}}{4}

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