If sin θ+2cos θ=1 ; then 2sin θ−cos θ=?


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Question Number 124309 by bemath last updated on 02/Dec/20

$I f \sin θ + 2 \cos θ = 1; t h e n$ $2 \sin θ - \cos θ = ?$
	Answered by MJS_new last updated on 02/Dec/20

	$t = \tan θ$ $\sin θ + 2 \cos θ = \frac{t + 2}{\sqrt{t^{2} + 1}} = 1 \Rightarrow t = - \frac{3}{4}$ $2 \sin θ - \cos θ = \frac{2 t - 1}{\sqrt{t^{2} + 1}} = - 2$
	Answered by liberty last updated on 02/Dec/20

	$l e t 2 \sin θ - \cos θ = p$ $\Leftrightarrow 4 \sin^{2} θ - 4 \sin θ \cos θ + \cos^{2} θ = p^{2}$ $\Leftrightarrow \sin^{2} θ + 4 \sin θ \cos θ + 4 \cos^{2} θ = 1$ $(1) + (2) \Rightarrow 5 = p^{2} + 1; g i v e p = \pm 2$ $H e n c e 2 \sin θ - \cos θ = \pm 2$
	Answered by Dwaipayan Shikari last updated on 02/Dec/20

	$t + 2 \sqrt{1 - t^{2}} = 1$ $4 - 4 t^{2} = 1 + t^{2} - 2 t \Rightarrow 5 t^{2} - 2 t - 3 = 0 \Rightarrow t = \frac{2 \pm 8}{10} . = 1, \frac{- 3}{5}$ $2 s i n θ - c o s θ = 2$
	Commented by bemath last updated on 02/Dec/20
	$sir liberty . your correct but if sin θ=−(3/5) then cos θ=(4/5) ( 4^(th) quadrant) and { ((2sin θ−cos θ=−(6/5)−(4/5)=−2)),((sin θ+2cos θ=−(3/5)+(8/5)=1)) :} Thanks you sir liberty and sir dwaipayan$
	$s i r l i b e r t y . y o u r c o r r e c t b u t$ $i f \sin θ = - \frac{3}{5} t h e n \cos θ = \frac{4}{5} (4^{t h} q u a d r a n t)$ $a n d {\begin{cases} 2 \sin θ - \cos θ = - \frac{6}{5} - \frac{4}{5} = - 2 \\ \sin θ + 2 \cos θ = - \frac{3}{5} + \frac{8}{5} = 1 \end{cases}$ $T h a n k s y o u s i r l i b e r t y a n d s i r d w a i p a y a n$
	Commented by liberty last updated on 02/Dec/20

	$i f \sin θ = - \frac{3}{5} \land \cos θ = - \frac{4}{5} t h e n$ $2 \sin θ - \cos θ = - \frac{6}{5} - (- \frac{4}{5}) = - \frac{2}{5}$ $a n d \sin θ + 2 \cos θ = - \frac{3}{5} + 2 (- \frac{4}{5}) = - \frac{3}{5} - \frac{8}{5} \neq 1$
	Commented by Dwaipayan Shikari last updated on 02/Dec/20

	$O h! t h a n k i n g y o u$
	Answered by Dwaipayan Shikari last updated on 02/Dec/20

	$s i n θ + 2 c o s θ = 1$ ${s i n}^{2} θ + 4 {c o s}^{2} θ + 4 s i n θ c o s θ = 1 \Rightarrow 4 {s i n}^{2} θ + {c o s}^{2} θ - 4 s i n θ c o s θ = 4$ ${(2 s i n θ - c o s θ)}^{2} = 4 \Rightarrow 2 s i n θ - c o s θ = \pm 2$
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