Trigonometric equations

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Let k be sum of all x in the interval $[0, 2 π]$ such that $3 \cot^{2} x + 8 \cot x + 3 = 0$ then the value of $\frac{k}{π}$ is

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Solution

$\begin{aligned} 3 \cot^{2} x + 8 \cot x + 3 = 0 \\ ∴ \cot x = \frac{- 4 \pm \sqrt{7}}{3} \end{aligned}$
Both roots are negative.
$\begin{array}{ll} ∴ \frac{π}{2} < x_{1}, x_{2} < π \\ ∴ π < x_{1} + x_{2} < 2 π \end{array}$
Product of roots $= \cot x_{1} \cot x_{2} = 1$
$\begin{array}{l} \Rightarrow \cot x_{1} \cot (\frac{3 π}{2} - x_{1}) = 1 \\ \Rightarrow x_{1} + x_{2} = \frac{3 π}{2} \end{array}$
Similarly, $x_{1} + x_{2} = \frac{7 π}{2}$

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Trigonometric equations

Remember concepts with our Masterclasses.

Let k be sum of all x in the interval [0,2π] such that 3cot2⁡x+8cot⁡x+3=0 then the value of kπ is

3cot2⁡x+8cot⁡x+3=0∴ cot⁡x=−4±73Both roots are negative.∴ π2<x1,x2<π∴ π<x1+x2<2πProduct of roots =cot⁡x1cot⁡x2=1⇒ cot⁡x1cot⁡3π2−x1=1⇒ x1+x2=3π2Similarly, x1+x2=7π2

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Let k be sum of all x in the interval $[0, 2 π]$ such that $3 \cot^{2} x + 8 \cot x + 3 = 0$ then the value of $\frac{k}{π}$ is