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Question Number 140187 by mathsuji last updated on 05/May/21
Solution equation:  sin2x=1+(√2) cosx+cos2x
$${Solution}\:{equation}: \\ $$$${sin}\mathrm{2}{x}=\mathrm{1}+\sqrt{\mathrm{2}}\:{cosx}+{cos}\mathrm{2}{x} \\ $$
Answered by Ankushkumarparcha last updated on 05/May/21
Solution: sin(2x) = 2cos^2 (x)+(√2)cos(x) (∵ cos(2x) = 2cos^2 (x)−1)  sin(x) = cos(x)+1/(√2)      (∵ sin(2x) = 2 sin(x)cos (x))  sin((Π/4) − x) = 1/2 => x = 2nΠ + ((5Π)/(12)) ,n∈Z
$${Solution}:\:\mathrm{sin}\left(\mathrm{2}{x}\right)\:=\:\mathrm{2cos}^{\mathrm{2}} \left({x}\right)+\sqrt{\mathrm{2}}\mathrm{cos}\left({x}\right)\:\left(\because\:\mathrm{cos}\left(\mathrm{2}{x}\right)\:=\:\mathrm{2cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1}\right) \\ $$$$\mathrm{sin}\left({x}\right)\:=\:\mathrm{cos}\left({x}\right)+\mathrm{1}/\sqrt{\mathrm{2}}\:\:\:\:\:\:\left(\because\:\mathrm{sin}\left(\mathrm{2}{x}\right)\:=\:\mathrm{2}\:\mathrm{sin}\left({x}\right)\mathrm{cos}\:\left({x}\right)\right) \\ $$$$\mathrm{sin}\left(\frac{\Pi}{\mathrm{4}}\:−\:{x}\right)\:=\:\mathrm{1}/\mathrm{2}\:=>\:{x}\:=\:\mathrm{2}{n}\Pi\:+\:\frac{\mathrm{5}\Pi}{\mathrm{12}}\:,{n}\in\mathbb{Z} \\ $$
Commented by mr W last updated on 05/May/21
or  cos x=0  ⇒x=nπ+(π/2)
$${or} \\ $$$$\mathrm{cos}\:{x}=\mathrm{0} \\ $$$$\Rightarrow{x}={n}\pi+\frac{\pi}{\mathrm{2}} \\ $$
Commented by mathsuji last updated on 06/May/21
thankyou sir
$${thankyou}\:{sir} \\ $$

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