The values of ‘b’ such that the equation $\frac{b \cos x}{2 \cos 2 x - 1} = \frac{b + \sin x}{(\cos^{2} x - 3 \sin^{2} x) \tan x}$ possess solutions, belong to the set

Question

The values of ‘b’ such that the equation $$bcosx2cos2x$$−$$1=b+sinxcos2x$$−$$3sin2xtanxpossess$$ solutions, belong to the set

Infinity Learn · Accepted Answer

Let us find domain of given equation $$i)2cos2x$$−$$1$$≠$$0$$⇒x≠nπ±π$$6ii)tanx$$≠$$0$$⇒x≠n$$π$$$$iii)cos2x$$−$$3sin2x$$≠$$0$$⇒x≠nπ±π$$6Also$$, $$2cos2x$$−$$1=2cos2x$$−$$sin2x$$−$$cos2x+sin2x=cos2x$$−$$3sin2xbcosxcos2x$$−$$3sin2x=b+sinxcos2x$$−$$3sin2xtanxbcosx=b+sinxsinxcosx$$ $$bsinx=b+sinx$$⇒$$sinx=bb$$−$$1Since$$ −$$1$$≤sinx≤$$1$$⇒$$-1$$≤$$bb-1$$≤$$1$$ ⇒$$bb-1+1$$≥$$0$$ and $$bb-1-1$$≤$$0$$ ⇒$$2b-1b-1$$≥$$0$$ and $$1b-1$$≤o ⇒b≤$$12$$ or b>$$1$$ and b<$$1$$⇒b<$$12$$ or b>$$1$$

The values of ‘b’ such that the equation $\frac{b \cos x}{2 \cos 2 x - 1} = \frac{b + \sin x}{(\cos^{2} x - 3 \sin^{2} x) \tan x}$ possess solutions, belong to the set

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The values of ‘b’ such that the equation bcosx2cos2x−1=b+sinxcos2x−3sin2xtanxpossess solutions, belong to the set

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The values of ‘b’ such that the equation $\frac{b \cos x}{2 \cos 2 x - 1} = \frac{b + \sin x}{(\cos^{2} x - 3 \sin^{2} x) \tan x}$ possess solutions, belong to the set