Question Number 192987 by Mingma last updated on 01/Jun/23
Answered by ajfour last updated on 01/Jun/23
$${let}\:{left}\:{vertical}=\mathrm{2} \\ $$$$\mathrm{2cos}\:\theta={c}\mathrm{cos}\:{x} \\ $$$$\mathrm{sin}\:\theta={c}\mathrm{sin}\:{x} \\ $$$$\mathrm{4cos}\:\theta\mathrm{cos}\:\theta={c}\mathrm{cos}\:\left(\pi−{x}−\theta\right) \\ $$$$……….\:\:\:\:\:\:\:\:………..\:\:\:\:\:\:\:……….. \\ $$$$\mathrm{4cos}\:^{\mathrm{2}} \theta+{c}\mathrm{cos}\:{x}\mathrm{cos}\:\theta={c}\mathrm{sin}\:{x}\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:^{\mathrm{2}} {x}+\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{2}}=\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)=\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\mathrm{sin}\:{x}=\sqrt{\frac{\mathrm{3}}{\mathrm{5}}} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$ \\ $$
Commented by Mingma last updated on 01/Jun/23
Perfect ��