for-what-value-of-x-0-lt-x-lt-2-4cosec-2-x-9-cotx-

Question Number 98191 by behi83417@gmail.com last updated on 12/Jun/20

for what value of : {x ∣ 0 < x < 2 π} :

4 {cosec}^{2} x - 9 = cotx

Answered by MJS last updated on 12/Jun/20

{4 cosec}^{2} x - 9 = \cot x

{9 \sin}^{2} x + \sin x \cos x - 4 = 0

9 \cos 2 x - \sin 2 x - 1 = 0

x = \arctan t

- \frac{10 t^{2} + 2 t - 8}{t^{2} + 1} = 0

t^{2} + \frac{1}{5} t - \frac{4}{5} = 0

t_{1} = - 1; t_{2} = \frac{4}{5}

\Rightarrow

x_{1.1} = \frac{3 π}{4}; x_{1.2} = \frac{7 π}{4}; x_{2.1} = \arctan \frac{4}{5}; x_{2.2} = π + \arctan \frac{4}{5}

Answered by 1549442205 last updated on 12/Jun/20

we have \frac{4}{\sin^{2} x} - 9 = \frac{cosx}{sinx} \Leftrightarrow 4 (1 + \cot^{2} x) - 9 = cotx

{4 \cot}^{2} x - cotx - 5 = 0 \Leftrightarrow (cotx + 1) (4 cotx - 5) = 0

a / cotx + 1 = 0 \Leftrightarrow \cot x = - 1 = \cot \frac{3 π}{4} \Rightarrow x \in {\frac{3 π}{4}; \frac{7 π}{4}}

a / 4 cotx - 5 = 0 \Leftrightarrow cotx = \frac{5}{4} \Leftrightarrow tanx = \frac{4}{5}

\Rightarrow x \in {\arctan \frac{4}{5}; \arctan \frac{4}{5} + π}

Thus, the roots in interval [0; 2 π] of

given equation are : x \in {\frac{3 π}{4}; \frac{7 π}{4}; \arctan \frac{4}{5}; \arctan \frac{4}{5} + π}

Commented by behi83417@gmail.com last updated on 12/Jun/20

thank you very much sir .