Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 167220 by tabata last updated on 09/Mar/22

Answered by MikeH last updated on 10/Mar/22

∣3x−4∣ ≥ 5 ⇒ 3x−4 ≤ −5 or 3x−4 ≥ 5 ⇒ x ≤−(1/3) or x ≥ 3

$$\mid\mathrm{3}{x}−\mathrm{4}\mid\:\geqslant\:\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{3}{x}−\mathrm{4}\:\leqslant\:−\mathrm{5}\:\mathrm{or}\:\mathrm{3}{x}−\mathrm{4}\:\geqslant\:\mathrm{5} \\ $$$$\Rightarrow\:{x}\:\leqslant−\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{or}\:{x}\:\geqslant\:\mathrm{3} \\ $$

Answered by MikeH last updated on 10/Mar/22

(1) sin θ = (1/2) cos^2 θ + ((1/2))^2 = 1 ⇒ cos^2 θ = (3/4) ⇒ cos θ = ((√3)/2) hence tan θ = (1/( (√3))) csc θ = (1/(sin θ)) = (1/((1/2))) = 2 sec θ = (1/(cos θ)) = (1/(((√3) /2))) = (2/( (√3))) cot θ = (1/(tan θ)) = (√3)

$$\left(\mathrm{1}\right)\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:\mathrm{cos}^{\mathrm{2}} \theta\:+\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\mathrm{1}\:\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \theta\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{cos}\:\theta\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\mathrm{hence}\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{csc}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}/\mathrm{2}\right)}\:=\:\mathrm{2} \\ $$$$\mathrm{sec}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\:=\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{3}}\:/\mathrm{2}\right)}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cot}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\:=\:\sqrt{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com