Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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)

$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{2}} {x}\:<\:\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x} \\ $$ $$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\right)\:<\:\mathrm{1} \\ $$ $$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x}\:<\:\mathrm{0} \\ $$ $${we}\:{get}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\:<\:{x}\:<\:\frac{\mathrm{4}\pi}{\mathrm{8}}\:\rightarrow{then}\:{a}+{b}\:=\:\frac{\mathrm{7}\pi}{\mathrm{8}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com