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Question Number 159049 by mnjuly1970 last updated on 12/Nov/21
        solve the following equation:     sin^( 6) (x) +cos^( 6) (x)+sin^4 (x)+cos^( 4) (x)=(3/4)
$$ \\ $$$$\:\:\:\:\:\:{solve}\:{the}\:{following}\:{equation}: \\ $$$$\: \\ $$$${sin}^{\:\mathrm{6}} \left({x}\right)\:+{cos}^{\:\mathrm{6}} \left({x}\right)+{sin}^{\mathrm{4}} \left({x}\right)+{cos}^{\:\mathrm{4}} \left({x}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$
Answered by gsk2684 last updated on 12/Nov/21
⇒(sin^2 x+cos^2 x)^3 −3 sin^2 x cos^2 x(sin^2 x+cos^2 x)  +(sin^2 x+cos^2 x)^2 −2 sin^2 x cos^2 x=(3/4)  ⇒1−3sin^2 x cos^2 x+1−2sin^2 x cos^2 x=(3/4)  ⇒(5/4)=5 sin^2 x cos^2 x  ⇒1=sin^2 2x  ∴ 2x=(2n+1)(π/2),n∈Z    x=(2n+1)(π/4),n∈Z        .....gsk....
$$\Rightarrow\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} −\mathrm{3}\:\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right) \\ $$$$+\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}\:\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{3sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{5}}{\mathrm{4}}=\mathrm{5}\:\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$$$\Rightarrow\mathrm{1}=\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\therefore\:\mathrm{2}{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}},{n}\in{Z} \\ $$$$\:\:{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}},{n}\in{Z} \\ $$$$\:\:\:\:\:\:…..{gsk}…. \\ $$
Commented by mnjuly1970 last updated on 12/Nov/21
thanks alot sir g.s.k
$${thanks}\:{alot}\:{sir}\:{g}.{s}.{k} \\ $$
Commented by gsk2684 last updated on 13/Nov/21
welcome
$${welcome} \\ $$
Answered by MJS_new last updated on 12/Nov/21
sin x =s ∧ cos x =(√(1−s^2 ))  ⇒  s^4 −s^2 +(1/4)=0  (s^2 −(1/2))^2 =0  ⇒  sin x =±((√2)/2)  ⇒  x=−(π/4)+((nπ)/2)
$$\mathrm{sin}\:{x}\:={s}\:\wedge\:\mathrm{cos}\:{x}\:=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${s}^{\mathrm{4}} −{s}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({s}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:{x}\:=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}=−\frac{\pi}{\mathrm{4}}+\frac{{n}\pi}{\mathrm{2}} \\ $$

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