Menu Close

If-x-0-pi-2-and-2cos-x-sin-x-cos-x-tan-2-x-lt-sec-2-x-has-solution-set-is-a-lt-x-lt-b-find-the-value-of-a-b-




Question Number 115348 by bemath last updated on 25/Sep/20
If x ∈ (0,(π/2)) and 2cos x(sin x+cos x)+tan^2 x < sec^2 x   has solution set is a<x<b. find the  value of a+b
$${If}\:{x}\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right)\:{and}\:\mathrm{2cos}\:{x}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)+\mathrm{tan}\:^{\mathrm{2}} {x}\:<\:\mathrm{sec}\:^{\mathrm{2}} {x}\: \\ $$$${has}\:{solution}\:{set}\:{is}\:{a}<{x}<{b}.\:{find}\:{the} \\ $$$${value}\:{of}\:{a}+{b} \\ $$
Answered by bobhans last updated on 25/Sep/20
⇒sin 2x+2cos^2 x+tan^2 x < 1+tan^2 x  ⇒sin 2x+2((1/2)+(1/2)cos 2x) < 1  ⇒sin 2x+cos 2x < 0  we get ((3π)/8) < x < ((4π)/8) →then a+b = ((7π)/8)
$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{2}} {x}\:<\:\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right)\:<\:\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x}\:<\:\mathrm{0} \\ $$$${we}\:{get}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\:<\:{x}\:<\:\frac{\mathrm{4}\pi}{\mathrm{8}}\:\rightarrow{then}\:{a}+{b}\:=\:\frac{\mathrm{7}\pi}{\mathrm{8}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *