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Question Number 30087 by rahul 19 last updated on 16/Feb/18

solve: cos3x.cos^3 x+sin 3x.sin^3 x=0.

$$\mathrm{solve}: \\ $$$$\mathrm{cos3}{x}.{cos}^{\mathrm{3}} {x}+\mathrm{sin}\:\mathrm{3}{x}.\mathrm{sin}\:^{\mathrm{3}} {x}=\mathrm{0}. \\ $$

Answered by MJS last updated on 16/Feb/18

x=(π/4)+((π∙z)/2); z∈Z sin(π/4)=cos(π/4)∧sin((3π)/4)=−cos((3π)/4) you can easily see this looking at the angles in a circle

$${x}=\frac{\pi}{\mathrm{4}}+\frac{\pi\centerdot{z}}{\mathrm{2}};\:{z}\in\mathbb{Z} \\ $$$$\mathrm{sin}\frac{\pi}{\mathrm{4}}=\mathrm{cos}\frac{\pi}{\mathrm{4}}\wedge\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{4}}=−\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{see}\:\mathrm{this}\:\mathrm{looking} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle} \\ $$

Answered by MJS last updated on 16/Feb/18

...the hard way: let me write c for cos and s for sin c(α+β)=c(α)c(β)−s(α)s(β) β=2α c(3α)=c(α)c(2α)−s(α)s(2α) c(2α)=2c^2 (α)−1 s(2α)=2s(α)c(α) c(3α)=2c^3 (α)−c(α)−2c(α)s^2 (α) s^2 (α)=1−c^2 (α) c(3α)=4c^3 (α)−3c(α) s(α+β)=s(α)c(β)+c(α)s(β) β=2α s(3α)=c(α)s(2α)+c(2α)s(α) using c(2α), s(2α) as above s(3α)=(4c^2 (α)−1)s(α) c(3α)c^3 (α)+s(3α)s^3 (α)=0 4c^4 (α)−3c^2 (α)+(4c^2 (α)−1)s^4 (α) s^4 (α)=(1−c^2 (α))^2 8c^6 (α)−12c^4 (α)+6c^2 (α)−1=0 c(α)=(√t) 8t^3 −12t^2 +6t−1=0 t^3 −(3/2)t^2 +(3/4)t−(1/8)=0 here you need a good eye, the left side is (t−(1/2))^3 so t_1 =t_2 =t_3 =(1/2) cos(α)=((2(√2))/2) α=(π/4)+((πz)/2); z∈Z

$$...\mathrm{the}\:\mathrm{hard}\:\mathrm{way}: \\ $$$$\mathrm{let}\:\mathrm{me}\:\mathrm{write}\:\mathrm{c}\:\mathrm{for}\:\mathrm{cos}\:\mathrm{and}\:\mathrm{s}\:\mathrm{for}\:\mathrm{sin} \\ $$$$\mathrm{c}\left(\alpha+\beta\right)=\mathrm{c}\left(\alpha\right)\mathrm{c}\left(\beta\right)−\mathrm{s}\left(\alpha\right)\mathrm{s}\left(\beta\right) \\ $$$$\beta=\mathrm{2}\alpha \\ $$$$\mathrm{c}\left(\mathrm{3}\alpha\right)=\mathrm{c}\left(\alpha\right)\mathrm{c}\left(\mathrm{2}\alpha\right)−\mathrm{s}\left(\alpha\right)\mathrm{s}\left(\mathrm{2}\alpha\right) \\ $$$$\mathrm{c}\left(\mathrm{2}\alpha\right)=\mathrm{2c}^{\mathrm{2}} \left(\alpha\right)−\mathrm{1} \\ $$$$\mathrm{s}\left(\mathrm{2}\alpha\right)=\mathrm{2s}\left(\alpha\right)\mathrm{c}\left(\alpha\right) \\ $$$$\mathrm{c}\left(\mathrm{3}\alpha\right)=\mathrm{2c}^{\mathrm{3}} \left(\alpha\right)−\mathrm{c}\left(\alpha\right)−\mathrm{2c}\left(\alpha\right)\mathrm{s}^{\mathrm{2}} \left(\alpha\right) \\ $$$$\mathrm{s}^{\mathrm{2}} \left(\alpha\right)=\mathrm{1}−\mathrm{c}^{\mathrm{2}} \left(\alpha\right) \\ $$$$\mathrm{c}\left(\mathrm{3}\alpha\right)=\mathrm{4c}^{\mathrm{3}} \left(\alpha\right)−\mathrm{3c}\left(\alpha\right) \\ $$$$ \\ $$$$\mathrm{s}\left(\alpha+\beta\right)=\mathrm{s}\left(\alpha\right)\mathrm{c}\left(\beta\right)+\mathrm{c}\left(\alpha\right)\mathrm{s}\left(\beta\right) \\ $$$$\beta=\mathrm{2}\alpha \\ $$$$\mathrm{s}\left(\mathrm{3}\alpha\right)=\mathrm{c}\left(\alpha\right)\mathrm{s}\left(\mathrm{2}\alpha\right)+\mathrm{c}\left(\mathrm{2}\alpha\right)\mathrm{s}\left(\alpha\right) \\ $$$$\mathrm{using}\:\mathrm{c}\left(\mathrm{2}\alpha\right),\:\mathrm{s}\left(\mathrm{2}\alpha\right)\:\mathrm{as}\:\mathrm{above} \\ $$$$\mathrm{s}\left(\mathrm{3}\alpha\right)=\left(\mathrm{4c}^{\mathrm{2}} \left(\alpha\right)−\mathrm{1}\right)\mathrm{s}\left(\alpha\right) \\ $$$$ \\ $$$$\mathrm{c}\left(\mathrm{3}\alpha\right)\mathrm{c}^{\mathrm{3}} \left(\alpha\right)+\mathrm{s}\left(\mathrm{3}\alpha\right)\mathrm{s}^{\mathrm{3}} \left(\alpha\right)=\mathrm{0} \\ $$$$\mathrm{4c}^{\mathrm{4}} \left(\alpha\right)−\mathrm{3c}^{\mathrm{2}} \left(\alpha\right)+\left(\mathrm{4c}^{\mathrm{2}} \left(\alpha\right)−\mathrm{1}\right)\mathrm{s}^{\mathrm{4}} \left(\alpha\right) \\ $$$$\mathrm{s}^{\mathrm{4}} \left(\alpha\right)=\left(\mathrm{1}−\mathrm{c}^{\mathrm{2}} \left(\alpha\right)\right)^{\mathrm{2}} \\ $$$$\mathrm{8c}^{\mathrm{6}} \left(\alpha\right)−\mathrm{12c}^{\mathrm{4}} \left(\alpha\right)+\mathrm{6c}^{\mathrm{2}} \left(\alpha\right)−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{c}\left(\alpha\right)=\sqrt{\mathrm{t}} \\ $$$$\mathrm{8t}^{\mathrm{3}} −\mathrm{12t}^{\mathrm{2}} +\mathrm{6t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{t}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\mathrm{t}−\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\mathrm{here}\:\mathrm{you}\:\mathrm{need}\:\mathrm{a}\:\mathrm{good}\:\mathrm{eye},\:\mathrm{the} \\ $$$$\mathrm{left}\:\mathrm{side}\:\mathrm{is} \\ $$$$\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\mathrm{so}\:\mathrm{t}_{\mathrm{1}} =\mathrm{t}_{\mathrm{2}} =\mathrm{t}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\left(\alpha\right)=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\alpha=\frac{\pi}{\mathrm{4}}+\frac{\pi{z}}{\mathrm{2}};\:{z}\in\mathbb{Z} \\ $$

Commented by rahul 19 last updated on 16/Feb/18

thanks.

$$\mathrm{thanks}. \\ $$

Answered by ajfour last updated on 16/Feb/18

(cos 3x)(4cos^3 x)+(sin 3x)(4sin^3 x)=0 cos 3x(cos 3x+3cos x)+ (sin 3x)(3sin x−sin 3x)=0 2cos^2 3x+6cos xcos 3x +6sin 3xsin x−2sin^2 3x =0 1+cos 6x+3(cos 4x+cos 2x) +3(cos 2x−cos 4x)−(1−cos 6x)=0 2cos 6x+6cos 2x =0 ⇒ 3cos 2x+cos 6x=0 or 3cos 2x+4cos^3 2x−3cos 2x=0 ⇒ cos^3 2x =0 or cos 2x =0 1−2sin^2 x=0 sin^2 x = (1/2) x = nπ±(π/4) .

$$\left(\mathrm{cos}\:\mathrm{3}{x}\right)\left(\mathrm{4cos}\:^{\mathrm{3}} {x}\right)+\left(\mathrm{sin}\:\mathrm{3}{x}\right)\left(\mathrm{4sin}\:^{\mathrm{3}} {x}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{cos}\:\mathrm{3}{x}\left(\mathrm{cos}\:\mathrm{3}{x}+\mathrm{3cos}\:{x}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:\mathrm{3}{x}\right)\left(\mathrm{3sin}\:{x}−\mathrm{sin}\:\mathrm{3}{x}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{2cos}\:^{\mathrm{2}} \mathrm{3}{x}+\mathrm{6cos}\:{x}\mathrm{cos}\:\mathrm{3}{x} \\ $$$$\:\:\:\:\:\:\:+\mathrm{6sin}\:\mathrm{3}{x}\mathrm{sin}\:{x}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{3}{x}\:=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{1}+\mathrm{cos}\:\mathrm{6}{x}+\mathrm{3}\left(\mathrm{cos}\:\mathrm{4}{x}+\mathrm{cos}\:\mathrm{2}{x}\right) \\ $$$$+\mathrm{3}\left(\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{4}{x}\right)−\left(\mathrm{1}−\mathrm{cos}\:\mathrm{6}{x}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{2cos}\:\mathrm{6}{x}+\mathrm{6cos}\:\mathrm{2}{x}\:=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{3cos}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{6}{x}=\mathrm{0} \\ $$$${or}\:\:\:\mathrm{3cos}\:\mathrm{2}{x}+\mathrm{4cos}\:^{\mathrm{3}} \mathrm{2}{x}−\mathrm{3cos}\:\mathrm{2}{x}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{2}{x}\:=\mathrm{0} \\ $$$${or}\:\:\:\:\:\:\mathrm{cos}\:\mathrm{2}{x}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{sin}\:^{\mathrm{2}} {x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{x}\:=\:{n}\pi\pm\frac{\pi}{\mathrm{4}}\:\:\:. \\ $$

Commented by rahul 19 last updated on 16/Feb/18

thank u sir.

$$\mathrm{thank}\:\mathrm{u}\:\mathrm{sir}. \\ $$

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