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Question Number 45842 by gunawan last updated on 17/Oct/18

Commented by maxmathsup by imad last updated on 17/Oct/18

$$whatsthislanguagesirjoel?$$

Commented by MJS last updated on 17/Oct/18

$$\mathrm{English}\mathrm{please}$$

Commented by Joel578 last updated on 17/Oct/18

$$\mathrm{It}\mathrm{used}\mathrm{Indonesian}\mathrm{language}$$

Commented by Joel578 last updated on 17/Oct/18

$$\mathrm{Translate}:$$$$\mathrm{Solve}\mathrm{the}\mathrm{equations}\mathrm{below}:$$$$a)$$$$b)$$$$c)$$$$d)$$$$e)$$$$f)$$$$g)$$$$\mathrm{Hint}:\mathrm{use}\mathrm{the}\mathrm{identity}a\cos x+b\sin x=k\cos (x-\alpha )$$

Commented by maxmathsup by imad last updated on 17/Oct/18

$$thankyou..$$

Commented by maxmathsup by imad last updated on 17/Oct/18

$$b)3cosx+sinx=3\iff \sqrt{10}(\frac{3}{\sqrt{10}}cosx+\frac{1}{\sqrt{10}}sinx)=3let$$$$cos\theta =\frac{3}{\sqrt{10}}andsin\theta =\frac{1}{\sqrt{10}}\Rightarrow tan\theta =\frac{1}{3}\Rightarrow \theta =arctan\left(\frac{1}{3}\right)$$$$\left(e\right)\Rightarrow \sqrt{10}cos(x-\theta )=3\Rightarrow cos(x-\theta )=\frac{3}{\sqrt{10}}=cos\theta \Rightarrow$$$$x-\theta =\theta +2k\pi orx-\theta =-\theta +2k\pi \left(kfromZ\right)\Rightarrow x=2\theta +2k\pi orx=2k\pi \Rightarrow$$$$x=2arctan\left(\frac{1}{3}\right)+2k\pi orx=2k\pi \left(kfromZ\right)$$$$$$

Commented by gunawan last updated on 18/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}$$

Commented by hknkrc46 last updated on 29/Nov/18

$$a)\cos x+\sqrt{3}\sin x=1$$$$\cos x+\sqrt{3}\sin x=1$$$$\cos x+\tan 60\sin x=1$$$$\cos x+\frac{\sin 60}{\cos 60}\sin x=1$$$$\frac{\cos x\cos 60+\sin x\sin 60}{\cos 60}=1$$$$\frac{\cos (x-60)}{\cos 60}=1\Rightarrow \cos (x-60)=\cos 60$$$$\Rightarrow x-60=60\vee x-60=-60$$$$\Rightarrow x=120\vee x=0$$$$b)3\cos x+\sin x=3$$$$3\cos x+\sin x=3\Rightarrow \sin x=3-3\cos x$$$$\Rightarrow \sin x=3(1-\cos x)$$$$\Rightarrow \sin x(1+\cos x)=3(1-\cos x)(1+\cos x)$$$$\Rightarrow \sin x(1+\cos x)=3(1-\cos {}^{2}x)$$$$\Rightarrow \sin x(1+\cos x)=3\sin {}^{2}x$$$$\Rightarrow 1+\cos x=3\sin x$$$$\Rightarrow 3\sin x-\cos x=1$$$$(3\cos x+\sin x)+3(3\sin x-\cos x)=10\sin x$$$$3+3\cdot 1=10\sin x\Rightarrow \sin x=\frac{6}{10}=\frac{3}{5}\Rightarrow x=36,86\approx 37$$$$x\approx 37\vee x\approx 143$$$$c)5\cos x+4\sin x=6$$$$\cos x=u\Rightarrow \sin x=\sqrt{1-u^2}$$$$5u+4\sqrt{1-u^2}=6\Rightarrow 4\sqrt{1-u^2}=6-5u$$$$\Rightarrow 16(1-u^2)={(6-5u)}^2$$$$\Rightarrow 16-16u^2=25u^2-60u+36$$$$\Rightarrow 41u^2-60u+20=0$$$$\Rightarrow u_{1,2}=\frac{60\pm \sqrt{{60}^2-4\cdot 41\cdot 20}}{82}=\frac{60\pm \sqrt{320}}{82}$$$$\approx \frac{60\pm 18}{82}\Rightarrow u_1=\cos x\approx \frac{39}{41}\vee u_2=\cos x\approx \frac{21}{41}$$$$x\approx {\cos }^{-1}\left(\frac{39}{41}\right)\vee x\approx {\cos }^{-1}\left(\frac{21}{41}\right)$$

Answered by Joel578 last updated on 17/Oct/18

$$\left(a\right)\cos x+\sqrt{3}\sin x=1$$$$$$$$k=\sqrt{1^2+{\left(\sqrt{3}\right)}^2}=2$$$$\tan \alpha =\frac{\sqrt{3}}{1}=\sqrt{3}\to \alpha =\frac{\pi }{3}$$$$$$$$\to 2\cos (x-\frac{\pi }{3})=1$$$$\cos (x-\frac{\pi }{3})=\frac{1}{2}$$$$x=\frac{2\pi }{3}+2n\pi \vee x=2n\pi (n\in \mathbb{Z})$$

Commented by gunawan last updated on 18/Oct/18

$$\mathrm{thanks}$$

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