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Question Number 100131 by bemath last updated on 25/Jun/20

$$\mathrm{If}3\sin \mathrm{x}+4\cos \mathrm{x}=5\mathrm{then}\sin \mathrm{x}=?$$

Commented by bobhans last updated on 25/Jun/20

$$\mathrm{let}\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t},\Rightarrow \sin \mathrm{x}=\frac{2\mathrm{t}}{1+{\mathrm{t}}^2},\cos \mathrm{x}=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}$$$$\iff \frac{6\mathrm{t}+4-{4\mathrm{t}}^2}{1+{\mathrm{t}}^2}=5\Rightarrow -{4\mathrm{t}}^2+6\mathrm{t}+4={5\mathrm{t}}^2+5$$$${9\mathrm{t}}^2-6\mathrm{t}+1=0,{(3\mathrm{t}-1)}^2=0$$$$\mathrm{t}=\frac{1}{3}\Rightarrow \sin \mathrm{x}=\frac{\frac{2}{3}}{1+\frac{1}{9}}=\frac{6}{10}=\frac{3}{5}$$

Commented by bemath last updated on 25/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{all}$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$${(3sinx+4cosx)}^2=25$$$$9{sin}^{2}x+24sinxcosx+16{cos}^{2}x=25({sin}^{2}x+{cos}^{2}x)$$$$16{sin}^{2}x-24sinxcosx+9{cos}^{2}x=0$$$${(4sinx-3cosx)}^2=0orcotx=\frac{4}{3}$$$$3+4cotx=\frac{5}{sinx}or\frac{sinx}{5}=\frac{3}{25}sosinx=\frac{3}{5}$$

Answered by Rasheed.Sindhi last updated on 25/Jun/20

$$\mathrm{If}3\sin \mathrm{x}+4\cos \mathrm{x}=5\mathrm{then}\sin \mathrm{x}=?$$$$4\cos \mathrm{x}=5-3\sin \mathrm{x}$$$${16\cos }^2\mathrm{x}={(5-3\sin \mathrm{x})}^2$$$$16(1-{\sin }^2\mathrm{x})=25-30\sin \mathrm{x}+{9\sin }^2\mathrm{x}$$$${25\sin }^2\mathrm{x}-30\sin \mathrm{x}+9=0$$$$\sin \mathrm{x}=\frac{30\pm \sqrt{900-900}}{50}=\frac{3}{5}$$

Answered by mr W last updated on 25/Jun/20

$$\cos x\cos \alpha +\sin x\sin \alpha =1$$$$with\sin \alpha =\frac{3}{5},\cos \alpha =\frac{4}{5}$$$$\cos (x-\alpha )=1$$$$x-\alpha =2k\pi$$$$x=2k\pi +\alpha$$$$\sin x=\sin \alpha =\frac{3}{5}$$

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