Question Number 70597 by Maclaurin Stickker last updated on 06/Oct/19
$${solve} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{1} \\ $$
Answered by Kunal12588 last updated on 08/Oct/19
$${cos}\:\mathrm{2}\alpha\:=\:\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} \:\alpha=\frac{\mathrm{1}+{cos}\:\mathrm{2}\alpha}{\mathrm{2}} \\ $$$${cos}^{\mathrm{2}} \beta=\frac{\mathrm{1}+{cos}\:\mathrm{2}\beta}{\mathrm{2}} \\ $$$${cos}^{\mathrm{2}} \mathrm{3}\beta=\frac{\mathrm{1}+{cos}\:\mathrm{6}\beta}{\mathrm{2}} \\ $$$$\therefore\:\mathrm{1}+{cos}\:\mathrm{2}\beta+\mathrm{1}+{cos}\:\mathrm{6}\beta=\mathrm{2} \\ $$$$\Rightarrow{cos}\:\mathrm{2}\beta+{cos}\:\mathrm{6}\beta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{cos}\:\mathrm{4}\beta\:{cos}\:\mathrm{2}\beta\:=\mathrm{0} \\ $$$$\Rightarrow{cos}\:\mathrm{4}\beta=\mathrm{0}\:\:\:{and}\:{cos}\:\mathrm{2}\beta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\beta=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{2}}\:\:\:\:{and}\:\mathrm{2}\beta=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\beta=\frac{{n}\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{8}},\:{n}\pi\pm\frac{\pi}{\mathrm{4}} \\ $$
Commented by mr W last updated on 06/Oct/19
$${nice}\:{way}! \\ $$$${better}\:{solution}\:{than}\:{mine}! \\ $$
Commented by mathmax by abdo last updated on 06/Oct/19
$${error}\:\:{cos}^{\mathrm{2}} \alpha=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\alpha\right)}{\mathrm{2}}\:.. \\ $$
Commented by Kunal12588 last updated on 08/Oct/19
$${thanks}\:{sir} \\ $$
Answered by mr W last updated on 06/Oct/19
$$\mathrm{cos}\:\mathrm{3}\beta=\mathrm{cos}\:\beta\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{3}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\beta+\mathrm{cos}^{\mathrm{2}} \:\beta\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${let}\:{t}=\mathrm{cos}^{\mathrm{2}} \:\beta\geqslant\mathrm{0},\:\leqslant\mathrm{1} \\ $$$$\Rightarrow{t}+{t}\left(\mathrm{4}{t}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{16}{t}^{\mathrm{3}} −\mathrm{24}{t}^{\mathrm{2}} +\mathrm{10}{t}−\mathrm{1}=\mathrm{0} \\ $$$${let}\:{s}=\mathrm{2}{t}\geqslant\mathrm{0},\leqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}{s}^{\mathrm{3}} −\mathrm{6}{s}^{\mathrm{2}} +\mathrm{5}{s}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({s}−\mathrm{1}\right)\left(\mathrm{2}{s}^{\mathrm{2}} −\mathrm{4}{s}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{1} \\ $$$$\Rightarrow{s}=\frac{\mathrm{2}\pm\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{t}=\mathrm{cos}^{\mathrm{2}} \:\beta=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{cos}\:\beta=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\beta={n}\pi\pm\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{t}=\mathrm{cos}^{\mathrm{2}} \:\beta=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}\:\Rightarrow\mathrm{cos}\:\beta=\pm\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\Rightarrow\beta={n}\pi\pm\frac{\pi}{\mathrm{8}} \\ $$$$\Rightarrow{t}=\mathrm{cos}^{\mathrm{2}} \:\beta=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{4}}\:\Rightarrow\mathrm{cos}\:\beta=\pm\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\Rightarrow\beta={n}\pi\pm\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$
Answered by Rasheed.Sindhi last updated on 06/Oct/19
$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{1} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{sin}^{\mathrm{2}} \beta+\mathrm{cos}^{\mathrm{2}} \beta\:\: \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{sin}^{\mathrm{2}} \beta \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta−\mathrm{sin}^{\mathrm{2}} \beta=\mathrm{0} \\ $$$$\left(\mathrm{cos}\:\mathrm{3}\beta−\mathrm{sin}\:\beta\right)\left(\mathrm{cos}\:\mathrm{3}\beta+\mathrm{sin}\:\beta\right)=\mathrm{0} \\ $$$$\left.\mathrm{cos}\:\mathrm{3}\beta−\mathrm{sin}\:\beta=\mathrm{0}\:\vee\:\mathrm{cos}\:\mathrm{3}\beta+\mathrm{sin}\:\beta\right) \\ $$$$\left.\mathrm{cos}\:\mathrm{3}\beta=\mathrm{sin}\:\beta\:\vee\:\mathrm{cos}\:\mathrm{3}\beta=−\mathrm{sin}\:\beta\right) \\ $$$$\mathrm{cos}\:\mathrm{3}\beta=\mathrm{sin}\:\beta\:\vee\:\mathrm{cos}\:\mathrm{3}\beta=\mathrm{sin}\:\left(−\beta\right) \\ $$$$\mathrm{3}\beta+\beta=\pi/\mathrm{2}\:\:\vee\:\mathrm{3}\beta−\beta=\pi/\mathrm{2} \\ $$$$\mathrm{4}\beta=\frac{\pi}{\mathrm{2}}\:\:\vee\:\mathrm{2}\beta=\frac{\pi}{\mathrm{2}} \\ $$$$\beta=\frac{\pi}{\mathrm{8}}\:,\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by ajfour last updated on 06/Oct/19
$$\mathrm{2}\beta=\theta \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \left(\theta−\beta\right)+\mathrm{cos}\:^{\mathrm{2}} \left(\theta+\beta\right)=\mathrm{1} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\theta−\mathrm{2}\beta\right)+\mathrm{cos}\:\left(\mathrm{2}\theta+\mathrm{2}\beta\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2}\theta+\mathrm{2}\beta\:=\:\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\left(\mathrm{2}\theta−\mathrm{2}\beta\right) \\ $$$$\Rightarrow\:\:\mathrm{6}\beta=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\mathrm{2}\beta \\ $$$$\Rightarrow\:\:\beta\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{8}}\:,\:\:\beta\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\beta\:=\:\frac{{k}\pi}{\mathrm{8}}\:. \\ $$