Solve the following equations: -3 2x+ 3x= -3 2x+ 3x - Vidyadip

Question

Solve the following equations: $\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$

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Answer

$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$ $(\sin\text{x}+\sin3\text{x})-3\sin2\text{x}-(\cos\text{x}+\cos3\text{x})+3\cos2\text{x}=0$ $2\sin2\text{x}\cos\text{x}-3\sin2\text{x}-2\cos2\text{x}\cos\text{x}+3\cos2\text{x}=0$ $\sin2\text{x}(2\cos\text{x}-3)-\cos2\text{x}(2\cos\text{x}-3)=0$ $(2\cos\text{x}-3)(\sin2\text{x}-\cos2\text{x})=0$ $\cos\text{x}=\frac{3}{2}$ or $\sin2\text{x}-\cos2\text{x}-\cos2\text{x}=0$ but $\cos\text{x}\in[-11]\Rightarrow\cos\text{x}\not=\frac{3}{2}$ $\sin2\text{x}=\cos2\text{x}$ $2\text{x}=\text{n}\pi+\frac{\pi}{4}$ $\text{x}=\frac{\text{n}\pi}{2}+\frac{\pi}{8}$

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