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Question Number 183563 by CrispyXYZ last updated on 27/Dec/22

$$\mathrm{repost}\mathrm{an}\mathrm{unsolved}\mathrm{question}\mathrm{Q}182552$$$$\mathrm{Find}\mathrm{the}\mathrm{period}\mathrm{of}\mathrm{the}\mathrm{following}:$$$$a\bullet \sin 4x\sin 3x$$$$b\bullet \sin \pi x+\cos x$$$$c\bullet \frac{2{\sin }^{2}3x-3\tan 4x+4\cot 6x}{\mid \mathrm{cosec}8x\mid -{\sec }^{3}10x+\sqrt{\cot 12x}}$$

Commented by Acem last updated on 27/Dec/22

$$Thanksfriendforremindingusthisquestion,$$$$hopefromallsharingtheirmethodes.$$

Answered by mr W last updated on 27/Dec/22

$$\left(a\right)$$$$f\left(x\right)=\sin 4x\sin 3x$$$$4T=2n\pi \Rightarrow T=\frac{3n\pi }{6}$$$$3T=2m\pi \Rightarrow T=\frac{4m\pi }{6}$$$$T_{min}=2\pi ✓$$$$\left(b\right)$$$$f\left(x\right)=\sin \pi x+\cos x$$$$\pi T=2n\pi \Rightarrow T=2n=eveninteger$$$$\Rightarrow T=2m\pi =eveninteger\times \pi$$$$\Rightarrow noperiodicfunction!$$$$\left(c\right)$$$$2{\sin }^{2}3x=1-\cos 6x$$$$6T=2a\pi \Rightarrow T=\frac{a\pi \times 20}{60}$$$$8T=2b\pi \Rightarrow T=\frac{b\pi \times 15}{60}$$$$10T=2c\pi \Rightarrow T=\frac{c\pi \times 12}{60}$$$$4T=d\pi \Rightarrow T=\frac{3d\pi }{12}$$$$6T=e\pi \Rightarrow T=\frac{2e\pi }{12}$$$$12T=f\pi \Rightarrow T=\frac{f\pi }{12}$$$$\Rightarrow T_{min}=\pi ✓$$

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