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Question Number 143964 by naka3546 last updated on 20/Jun/21

$$x_{1}and{x}_{2}aresolutionsofequality:$$$$\cos \left(\frac{\pi x+\pi }{6}\right)-\sin \left(\frac{\pi x-\pi }{6}\right)=\frac{1}{2}\sqrt{3},0⩽x⩽12$$$$Findthevalueof{x}_1+x_2.$$

Commented by Canebulok last updated on 20/Jun/21

$$$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$Bysquaringbothsides,$$$$\Rightarrow cos(\frac{\pi x}{6}+\frac{\pi }{6})-sin(\frac{\pi x}{6}-\frac{\pi }{6})=\frac{\sqrt{3}}{2}$$$$\Rightarrow [cos\left(\frac{\pi x}{6}\right)cos\left(\frac{\pi }{6}\right)-sin\left(\frac{\pi x}{6}\right)sin\left(\frac{\pi }{6}\right)]-[sin\left(\frac{\pi x}{6}\right)cos\left(\frac{\pi }{6}\right)-cos\left(\frac{\pi x}{6}\right)sin\left(\frac{\pi }{6}\right)]=\frac{\sqrt{3}}{2}$$

Commented by Canebulok last updated on 20/Jun/21

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$\Rightarrow [cos\left(\frac{\pi x}{6}\right)cos\left(\frac{\pi }{6}\right)-sin\left(\frac{\pi x}{6}\right)cos\left(\frac{\pi }{6}\right)]+[cos\left(\frac{\pi x}{6}\right)sin\left(\frac{\pi }{6}\right)-sin\left(\frac{\pi x}{6}\right)sin\left(\frac{\pi }{6}\right)]=\frac{\sqrt{3}}{2}$$$$\Rightarrow cos\left(\frac{\pi }{6}\right)[cos\left(\frac{\pi x}{6}\right)-sin\left(\frac{\pi x}{6}\right)]+sin\left(\frac{\pi }{6}\right)[cos\left(\frac{\pi x}{6}\right)-sin\left(\frac{\pi x}{6}\right)]=\frac{\sqrt{3}}{2}$$$$\Rightarrow [cos\left(\frac{\pi }{6}\right)+sin\left(\frac{\pi }{6}\right)][cos\left(\frac{\pi x}{6}\right)-sin\left(\frac{\pi x}{6}\right)]=\frac{\sqrt{3}}{2}$$$$$$$$Bysquaringbothsides,$$$$\Rightarrow [1+2cos\left(\frac{\pi }{6}\right)sin\left(\frac{\pi }{6}\right)][1-2sin\left(\frac{\pi x}{6}\right)cos\left(\frac{\pi x}{6}\right)]=\frac{3}{4}$$$$\Rightarrow [1+sin\left(\frac{\pi }{3}\right)][1-sin\left(\frac{\pi x}{3}\right)]=\frac{3}{4}$$$$\Rightarrow 1-sin\left(\frac{\pi x}{3}\right)=\frac{3}{4[1+sin\left(\frac{\pi }{3}\right)]}$$$$$$$$\Rightarrow -sin\left(\frac{\pi x}{3}\right)=\frac{3}{(4+4sin\left(\frac{\pi }{3}\right))}-1$$$$$$$$\Rightarrow sin\left(\frac{\pi x}{3}\right)=1-\frac{3}{(4+4sin\left(\frac{\pi }{3}\right))}$$$$\Rightarrow arcsin[1-\frac{3}{(4+4sin\left(\frac{\pi }{3}\right))}]=\frac{\pi x}{3}$$$$\Rightarrow arcsin[1-\frac{3}{(4+4sin\left(\frac{\pi }{3}\right))}]\left(\frac{3}{\pi }\right)=x$$$$$$$$\sim Kevin$$

Answered by bramlexs22 last updated on 20/Jun/21

$$\sin \left(\frac{\pi x-\pi }{6}\right)=\cos \left(\frac{3\pi -\pi x+\pi }{6}\right)=\cos \left(\frac{\pi x-4\pi }{6}\right)$$$$\iff \cos \left(\frac{\pi x+\pi }{6}\right)-\cos \left(\frac{\pi x-4\pi }{6}\right)=\frac{\sqrt{3}}{2}$$$$\iff -2\sin \left(\frac{2\pi x-3\pi }{12}\right)\sin \left(\frac{5\pi }{12}\right)=\frac{\sqrt{3}}{2}$$$$\iff$$

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