Let S be the sum of all x in the interval -0-2- such that 3 2 x - 8 x -3-0- then the value of S- is

3 ^2 x + 8 x +3=0 ⇒ x= 4 ±√(7)3 Both roots are negative. ∴π2 x_1,x_2 π or nbsp;3π2 x_1,x_2 2π ∴π x_1 +x_2 2 π or 3π x_1 +x_2 4π Product of roots = x_1 ...

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Standard XI

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

Let S be the ...

Question

Let $S$ be the 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 $\frac{S}{π}$ is

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Solution

The correct option is C $5$
$3 {cot}^{2} x + 8 cot x + 3 = 0$
$\Rightarrow cot x = \frac{- 4 \pm \sqrt{7}}{3}$
Both roots are negative.
$∴ \frac{π}{2} < x_{1}, x_{2} < π$ or $\frac{3 π}{2} < x_{1}, x_{2} < 2 π$
$∴ π < x_{1} + x_{2} < 2 π$ or $3 π < x_{1} + x_{2} < 4 π$

Product of roots $= cot x_{1} cot x_{2} = 1$
$\Rightarrow cot x_{1} cot (\frac{3 π}{2} - x_{1}) = 1$
$\Rightarrow x_{1} + x_{2} = \frac{3 π}{2}$
Similarly, $x_{1} + x_{2} = \frac{7 π}{2}$

S = $\frac{3 π}{2} + \frac{7 π}{2} = 5 π$

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Let S be the sum of all x in the interval [0,2π] such that 3cot2x+8cotx+3=0, then the value of Sπ is

The correct option is C 5 3cot2x+8cotx+3=0 ⇒cotx=−4±√73 Both roots are negative. ∴π2<x1,x2<π or 3π2<x1,x2<2π ∴π<x1+x2<2π or 3π<x1+x2<4π Product of roots =cotx1cotx2=1 ⇒cotx1cot(3π2−x1)=1 ⇒x1+x2=3π2 Similarly, x1+x2=7π2 S = 3π2+7π2=5π

Let $S$ be the 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 $\frac{S}{π}$ is