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Question Number 76622 by Sathvik last updated on 28/Dec/19

$$\mathrm{Find}(\mathrm{a}.\mathrm{b}.\mathrm{c})\mathrm{for}\mathrm{equation}\mathrm{acos}2\mathrm{x}+\mathrm{bsin}{}^2\mathrm{x}+\mathrm{c}=0\mathrm{is}\mathrm{satisfied}\mathrm{by}\mathrm{every}\mathrm{x}$$

Answered by MJS last updated on 28/Dec/19

$$a\cos 2x+b{\sin }^{2}x+c=0$$$$a\cos 2x+b\times \frac{1}{2}(1-\cos 2x)+c=0$$$$(a-\frac{b}{2})\cos 2x+\frac{b}{2}+c=0$$$$p\cos 2x+q=0\mathrm{\forall }x\in \mathbb{R}\Rightarrow p=q=0$$$$a-\frac{b}{2}=\frac{b}{2}+c\Rightarrow a=b+c$$$$a-\frac{b}{2}=0\to \frac{b}{2}+c=0\Rightarrow b=-2c\Rightarrow a=-c$$$$c=-t\in \mathbb{R}$$$$b=2t$$$$a=t$$$$\Rightarrow t\cos 2x+2t{\sin }^{2}x-t=0$$$$t(\cos 2x+{2\sin }^{2}x-1)=0$$$$\mathrm{but}\cos 2x+{2\sin }^{2}x-1=0$$$$\Rightarrow 0t=0\mathrm{true}\mathrm{for}\mathrm{all}t,x\in \mathbb{R}$$

Answered by john santu last updated on 29/Dec/19

$$a\cos 2x+b(\frac{1}{2}-\frac{1}{2}\cos 2x)=-c$$$$(a-\frac{1}{2}b)\cos 2x=-c-\frac{1}{2}b$$$$\cos 2x=\frac{-2c-b}{2a-b}\to -1⩽\frac{-2c-b}{2a-b}⩽1$$$$$$

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