Menu Close

Solve-for-x-in-the-range-0-x-2pi-the-equations-a-cos-x-pi-3-0-b-sin-x-cos-x-c-sin-2x-2sin-x-1-cos-x-

Tinku Tara 0 Comments
Pin




Question Number 39587 by Rio Mike last updated on 08/Jul/18
Solve for x in the range 0 ≤ x ≤2π the equations a) cos(x + (π/3)) = 0 b) sin x = cos x. c) sin 2x + 2sin x = 1 + cos x
$${Solve}\:{for}\:{x}\:{in}\:{the}\:{range}\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\mathrm{2}\pi \\ $$$${the}\:{equations} \\ $$$$\left.{a}\right)\:{cos}\left({x}\:+\:\frac{\pi}{\mathrm{3}}\right)\:=\:\mathrm{0}\: \\ $$$$\left.{b}\right)\:{sin}\:{x}\:=\:{cos}\:{x}. \\ $$$$\left.{c}\right)\:{sin}\:\mathrm{2}{x}\:+\:\mathrm{2}{sin}\:{x}\:=\:\mathrm{1}\:+\:{cos}\:{x} \\ $$$$ \\ $$
Answered by Joel579 last updated on 08/Jul/18
cos (x + (π/3)) = 0 x + (π/3) = nπ + (π/2) (n = 0, 1) x = (π/6) ∨ x = ((7π)/6)
$$\mathrm{cos}\:\left({x}\:+\:\frac{\pi}{\mathrm{3}}\right)\:=\:\mathrm{0} \\ $$$${x}\:+\:\frac{\pi}{\mathrm{3}}\:=\:{n}\pi\:+\:\frac{\pi}{\mathrm{2}}\:\:\:\:\left({n}\:=\:\mathrm{0},\:\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\pi}{\mathrm{6}}\:\:\vee\:\:\:{x}\:=\:\frac{\mathrm{7}\pi}{\mathrm{6}} \\ $$
Answered by Joel579 last updated on 08/Jul/18
sin x = cos x tan x = 1 x = nπ + (π/4) (n = 0) x = (π/4)
$$\mathrm{sin}\:{x}\:=\:\mathrm{cos}\:{x} \\ $$$$\mathrm{tan}\:{x}\:=\:\mathrm{1} \\ $$$${x}\:=\:{n}\pi\:+\:\frac{\pi}{\mathrm{4}}\:\:\left({n}\:=\:\mathrm{0}\right) \\ $$$${x}\:=\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by Joel579 last updated on 08/Jul/18
2 sin x cos x + 2sin x = 1 + cos x 2sin x(cos x + 1) = 1 + cos x 2sin x(1 + cos x) − (1 + cos x) = 0 (2sin x − 1)(1 + cos x) = 0 sin x = (1/2) → x = (π/6), ((5π)/6) cos x = −1 → x = π
$$\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:+\:\mathrm{2sin}\:{x}\:=\:\mathrm{1}\:+\:\mathrm{cos}\:{x} \\ $$$$\mathrm{2sin}\:{x}\left(\mathrm{cos}\:{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{1}\:+\:\mathrm{cos}\:{x} \\ $$$$\mathrm{2sin}\:{x}\left(\mathrm{1}\:+\:\mathrm{cos}\:{x}\right)\:−\:\left(\mathrm{1}\:+\:\mathrm{cos}\:{x}\right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2sin}\:{x}\:−\:\mathrm{1}\right)\left(\mathrm{1}\:+\:\mathrm{cos}\:{x}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\rightarrow\:\:{x}\:=\:\frac{\pi}{\mathrm{6}},\:\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$$$\mathrm{cos}\:{x}\:=\:−\mathrm{1}\:\:\rightarrow\:\:{x}\:=\:\pi \\ $$
Answered by MJS last updated on 08/Jul/18
(a) x+(π/3)=arccos 0 ={(π/2); ((3π)/2)} x={(π/2)−(π/3); ((3π)/2)−(π/3)}={(π/6); ((7π)/6)} (b) think of the circle x^2 +y^2 =1 again cos θ =sin θ means we′re looking for the points (((cos θ)),((sin θ)) ) = ((p),(q) ) with p=q ⇒ ⇒ p=q=((√2)/2) ∨ p=q=−((√2)/2) ⇒ ⇒ θ={−((5π)/4); (π/4)} (c) sin 2x=2sin x cos x 2sc+2s=1+c 2sc−c+2s−1=0 c(2s−1)+2s−1=0 (2s−1)(c+1)=0 2sin x −1=0 ∨ cos x +1=0 sin x =(1/2) ∨ cos x =−1 x=({(π/6); ((5π)/6)} ∪ {π})={(π/6); ((5π)/6); π}
$$\left({a}\right)\:{x}+\frac{\pi}{\mathrm{3}}=\mathrm{arccos}\:\mathrm{0}\:=\left\{\frac{\pi}{\mathrm{2}};\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right\} \\ $$$$\:\:\:\:\:{x}=\left\{\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}};\:\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right\}=\left\{\frac{\pi}{\mathrm{6}};\:\frac{\mathrm{7}\pi}{\mathrm{6}}\right\} \\ $$$$\left({b}\right)\:\mathrm{think}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\:\mathrm{again} \\ $$$$\:\:\:\:\:\mathrm{cos}\:\theta\:=\mathrm{sin}\:\theta\:\mathrm{means}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{the} \\ $$$$\:\:\:\:\:\mathrm{points}\:\begin{pmatrix}{\mathrm{cos}\:\theta}\\{\mathrm{sin}\:\theta}\end{pmatrix}\:=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\mathrm{with}\:{p}={q}\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:{p}={q}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\vee\:{p}={q}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\: \\ $$$$\:\:\:\:\:\Rightarrow\:\theta=\left\{−\frac{\mathrm{5}\pi}{\mathrm{4}};\:\frac{\pi}{\mathrm{4}}\right\} \\ $$$$\left({c}\right)\:\mathrm{sin}\:\mathrm{2}{x}=\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\mathrm{2}{sc}+\mathrm{2}{s}=\mathrm{1}+{c} \\ $$$$\:\:\:\:\:\mathrm{2}{sc}−{c}+\mathrm{2}{s}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:{c}\left(\mathrm{2}{s}−\mathrm{1}\right)+\mathrm{2}{s}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{s}−\mathrm{1}\right)\left({c}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{2sin}\:{x}\:−\mathrm{1}=\mathrm{0}\:\vee\:\mathrm{cos}\:{x}\:+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{sin}\:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\vee\:\mathrm{cos}\:{x}\:=−\mathrm{1} \\ $$$$\:\:\:\:\:{x}=\left(\left\{\frac{\pi}{\mathrm{6}};\:\frac{\mathrm{5}\pi}{\mathrm{6}}\right\}\:\cup\:\left\{\pi\right\}\right)=\left\{\frac{\pi}{\mathrm{6}};\:\frac{\mathrm{5}\pi}{\mathrm{6}};\:\pi\right\} \\ $$

Pin

Leave a Reply

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