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Question Number 73538 by Rio Michael last updated on 13/Nov/19

$$findthegeneralsolutionof$$$$sin4x+cos2x=0$$

Answered by ajfour last updated on 13/Nov/19

$$\cos 2x(2\sin 2x+1)=0$$$$\Rightarrow 2x=(2n+1)\frac{\pi }{2}or$$$$2x=(4n-1)\frac{\pi }{2}\pm \frac{\pi }{3}.$$

Commented by Rio Michael last updated on 13/Nov/19

$$thankssir$$

Answered by malwaan last updated on 14/Nov/19

$$2\mathit{s}\mathit{i}\mathit{n}2\mathit{x}\mathit{c}\mathit{o}\mathit{s}2\mathit{x}+\mathit{c}\mathit{o}\mathit{s}2\mathit{x}=0$$$$\mathit{c}\mathit{o}\mathit{s}2\mathit{x}(2\mathit{s}\mathit{i}\mathit{n}2\mathit{x}+1)=0$$$$\mathit{c}\mathit{o}\mathit{s}2\mathit{x}=0\Rightarrow 2\mathit{x}=\frac{\mathit{\pi }}{2}+2\mathit{n}\mathit{\pi }$$$$\Rightarrow \mathit{x}=\frac{\mathit{\pi }}{4}+\mathit{n}\mathit{\pi }$$$$\mathit{o}\mathit{r}2\mathit{s}\mathit{i}\mathit{n}2\mathit{x}+1=0$$$$\Rightarrow 2\mathit{s}\mathit{i}\mathit{n}2\mathit{x}=-1\Rightarrow \mathit{s}\mathit{i}\mathit{n}2\mathit{x}=-\frac{1}{2}$$$$\Rightarrow 2\mathit{x}=(\frac{\mathit{\pi }}{6}+\mathit{\pi })+2\mathit{n}\mathit{\pi }=\frac{7\mathit{\pi }}{6}+2\mathit{n}\mathit{\pi }$$$$\Rightarrow \mathit{x}=\frac{7\mathit{\pi }}{12}+\mathit{n}\mathit{\pi }$$$$\mathit{o}\mathit{r}2\mathit{x}=(2\mathit{\pi }-\frac{\mathit{\pi }}{6})+2\mathit{n}\mathit{\pi }$$$$=\frac{11\mathit{\pi }}{6}+2\mathit{n}\mathit{\pi }$$$$\Rightarrow \mathit{x}=\frac{11\mathit{\pi }}{12}+\mathit{n}\mathit{\pi }$$$$\Rightarrow \mathit{x}=\{(\frac{\mathit{\pi }}{4}\vee \frac{7\mathit{\pi }}{12}\vee \frac{11\mathit{\pi }}{12})+\mathit{n}\mathit{\pi }\}$$

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