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Question-167220




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

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