Menu Close

sin-2-x-cos-2-2x-sin-2-3x-3-2-

Tinku Tara 0 Comments
Pin




Question Number 203325 by depressiveshrek last updated on 16/Jan/24
sin^2 x+cos^2 (2x)+sin^2 (3x)=(3/2)
$$\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by esmaeil last updated on 16/Jan/24
sin^2 x+(1−2sin^2 x)^2 + (3sinx−4sin^3 x)^2 =(3/2)→ sin^2 x+1+4sin^4 x−4sin^2 x+ 9sin^2 x+16sin^6 x−24sin^4 x=(3/2)→ 16sin^6 x−20sin^4 x+6sin^2 x=(3/2)→ 32m^6 −40m^4 +12m^2 −3=0→ 4(2m^2 )^3 −5(2m^2 )^2 +6(2m^2 )−3=0 4p^3 −5p^2 +6p−3=0 4p^3 −4p^2 +4p−p^2 +2p−4+1=0 4(p^3 −p^2 +p)−(p−1)^2 −4=0 4(p^3 −p^2 +p−1)−(p−1)^2 =0 4(p−1)(p^2 +1)−(p−1)^2 =0 (p−1)(4p^2 +4−p+1)=0 p=1→2m^2 =1→m=±((√2)/2)→ sinx=sin(±(π/4))→ { ((x=2kπ±(π/4))),((x=2kπ+((3π)/4),x=2kπ+((5π)/4))) :} 4p^2 −p+5=0×
$${sin}^{\mathrm{2}} {x}+\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {x}\overset{\mathrm{2}} {\right)}+ \\ $$$$\left(\mathrm{3}{sinx}−\mathrm{4}{sin}^{\mathrm{3}} {x}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}\rightarrow \\ $$$${sin}^{\mathrm{2}} {x}+\mathrm{1}+\mathrm{4}{sin}^{\mathrm{4}} {x}−\mathrm{4}{sin}^{\mathrm{2}} {x}+ \\ $$$$\mathrm{9}{sin}^{\mathrm{2}} {x}+\mathrm{16}{sin}^{\mathrm{6}} {x}−\mathrm{24}{sin}^{\mathrm{4}} {x}=\frac{\mathrm{3}}{\mathrm{2}}\rightarrow \\ $$$$\mathrm{16}{sin}^{\mathrm{6}} {x}−\mathrm{20}{sin}^{\mathrm{4}} {x}+\mathrm{6}{sin}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{2}}\rightarrow \\ $$$$\mathrm{32}{m}^{\mathrm{6}} −\mathrm{40}{m}^{\mathrm{4}} +\mathrm{12}{m}^{\mathrm{2}} −\mathrm{3}=\mathrm{0}\rightarrow \\ $$$$ \\ $$$$\mathrm{4}\left(\mathrm{2}{m}^{\mathrm{2}} \right)^{\mathrm{3}} −\mathrm{5}\left(\mathrm{2}{m}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{6}\left(\mathrm{2}{m}^{\mathrm{2}} \right)−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{4}{p}^{\mathrm{3}} −\mathrm{5}{p}^{\mathrm{2}} +\mathrm{6}{p}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{4}{p}^{\mathrm{3}} −\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}{p}−{p}^{\mathrm{2}} +\mathrm{2}{p}−\mathrm{4}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{4}\left({p}^{\mathrm{3}} −{p}^{\mathrm{2}} +{p}\right)−\left({p}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{4}\left({p}^{\mathrm{3}} −{p}^{\mathrm{2}} +{p}−\mathrm{1}\right)−\left({p}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)−\left({p}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({p}−\mathrm{1}\right)\left(\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}−{p}+\mathrm{1}\right)=\mathrm{0} \\ $$$${p}=\mathrm{1}\rightarrow\mathrm{2}{m}^{\mathrm{2}} =\mathrm{1}\rightarrow{m}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\rightarrow \\ $$$${sinx}={sin}\left(\pm\frac{\pi}{\mathrm{4}}\right)\rightarrow\begin{cases}{{x}=\mathrm{2}{k}\pi\pm\frac{\pi}{\mathrm{4}}}\\{{x}=\mathrm{2}{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}},{x}=\mathrm{2}{k}\pi+\frac{\mathrm{5}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{4}{p}^{\mathrm{2}} −{p}+\mathrm{5}=\mathrm{0}× \\ $$$$ \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *