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Question Number 167986 by DAVONG last updated on 31/Mar/22

cosxcos(π/6)−sin(π/6)sinx=(π/4)

$$\mathrm{cosxcos}\frac{\pi}{\mathrm{6}}−\mathrm{sin}\frac{\pi}{\mathrm{6}}\mathrm{sinx}=\frac{\pi}{\mathrm{4}} \\ $$

Answered by Rasheed.Sindhi last updated on 31/Mar/22

cosxcos(π/6)−sin(π/6)sinx=(π/4) ⇒cos(x+(π/6))=(π/4) x+(π/6)=cos^(−1) ((π/4)) x=cos^(−1) ((π/4))−(π/6) • x=cos^(−1) ((π/4))+2nπ−(π/6) •x=cos^(−1) (2π−(π/4))+2nπ−(π/6) =cos^(−1) (((7π)/4))+2nπ−(π/6)

$$\mathrm{cosxcos}\frac{\pi}{\mathrm{6}}−\mathrm{sin}\frac{\pi}{\mathrm{6}}\mathrm{sinx}=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{cos}\left({x}+\frac{\pi}{\mathrm{6}}\right)=\frac{\pi}{\mathrm{4}}\: \\ $$$$\:\:\:\:\:{x}+\frac{\pi}{\mathrm{6}}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:{x}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{4}}\right)−\frac{\pi}{\mathrm{6}} \\ $$$$\:\:\:\bullet\:\:{x}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{4}}\right)+\mathrm{2}{n}\pi−\frac{\pi}{\mathrm{6}} \\ $$$$\:\:\:\:\bullet{x}=\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}\pi−\frac{\pi}{\mathrm{4}}\right)+\mathrm{2}{n}\pi−\frac{\pi}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{7}\pi}{\mathrm{4}}\right)+\mathrm{2}{n}\pi−\frac{\pi}{\mathrm{6}} \\ $$$$\:\:\:\:\: \\ $$

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