The value of a in (−π,0) satisfying sin⁡α+∫α2α cos⁡2xdx=0 is

sin⁡α+∫a2a cos⁡2xdx=0⇒sin⁡α+12(sin⁡4α−sin⁡2α)=0⇒sin⁡α[1+cos⁡(3α)]=0As −π<α<0,sin⁡α≠0 thereforecos⁡(3α)=−1⇒α=−π/3.

The value of a in (−π,0) satisfying sin⁡α+∫α2α cos⁡2xdx=0 is

sin⁡α+∫a2a cos⁡2xdx=0⇒sin⁡α+12(sin⁡4α−sin⁡2α)=0⇒sin⁡α[1+cos⁡(3α)]=0As −π<α<0,sin⁡α≠0 thereforecos⁡(3α)=−1⇒α=−π/3.

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The value of $a$ in $(- π, 0)$ satisfying $\sin α + \int_{α}^{2 α} \cos 2 x d x = 0$ is

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$- π / 2$

$- π$

$- π / 3$

$- π / 4$

answer is C.

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$\begin{array}{l} \sin α + \int_{a}^{2 a} \cos 2 x d x = 0 \\ \Rightarrow \sin α + \frac{1}{2} (\sin 4 α - \sin 2 α) = 0 \\ \Rightarrow \sin α [1 + \cos (3 α)] = 0 \end{array}$

As $- π < α < 0, \sin α \neq 0$ therefore

$\cos (3 α) = - 1 \Rightarrow α = - π / 3$ .

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The value of a in (−π,0) satisfying sin⁡α+∫α2α cos⁡2xdx=0 is

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The value of $a$ in $(- π, 0)$ satisfying $\sin α + \int_{α}^{2 α} \cos 2 x d x = 0$ is