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Question Number 107534 by bemath last updated on 11/Aug/20

⊨BeMath⊨ sin^6 x + cos^6 x = (7/(16)) ; x ∈ (0,(π/2))

$$\:\:\:\vDash\mathcal{B}{e}\mathcal{M}{ath}\vDash \\ $$$$\:\mathrm{sin}\:^{\mathrm{6}} {x}\:+\:\mathrm{cos}\:^{\mathrm{6}} {x}\:=\:\frac{\mathrm{7}}{\mathrm{16}}\:;\:{x}\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$

Commented by hgrocks last updated on 11/Aug/20

7 and 16 both my favorite numbers ����

Commented by bemath last updated on 11/Aug/20

thank you all sir

$${thank}\:{you}\:{all}\:{sir} \\ $$

Answered by Rio Michael last updated on 11/Aug/20

(sin^2 x + cos^2 x)^3 = (sin^2 x + cos^2 x)(sin^2 x + cos^2 x)(sin^2 x + cos^2 x) = (sin^4 x + 2sin^2 x cos^2 x + cos^4 x)(sin^2 x + cos^2 x) = sin^6 x + sin^4 x cos^2 x + 2sin^4 x cos^2 x + 2sin^2 x cos^4 x + sin^2 x cos^4 x + cos^6 x = sin^6 x + cos^6 x + 3sin^4 xcos^2 x + 3 sin^2 x cos^4 x ⇒ 1 − 3 sin^4 x cos^2 x −3 sin^2 x cos^4 x = sin^6 x + cos^6 x ⇒ sin^6 x + cos^6 x = 1−3sin^2 x cos^2 x( sin^2 x + cos^2 x) = 1− (3/4) sin^2 2x ⇒1 − (3/4) sin^2 2x = (7/(16)) ; x ∈(0,(π/2)) −(3/4) sin^2 2x = −(9/(16)) or sin^2 2x = (3/4) ⇒ sin 2x = ((√3)/2) or 2x = πn + (−1)^n (π/3) x = ((πn)/2) + (−1)^n (π/6) n = 0, x = (π/6) n = 1, x = (π/3) n = 2, x = out of range given ⇒ solution set S = {(π/6),(π/3)}

$$\left(\mathrm{sin}^{\mathrm{2}} \:{x}\:+\:\mathrm{cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}} \:=\:\left(\mathrm{sin}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{2}} {x}\right)\left(\mathrm{sin}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{2}} {x}\right)\left(\mathrm{sin}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{2}} {x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{sin}^{\mathrm{4}} {x}\:+\:\mathrm{2sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{4}} {x}\right)\left(\mathrm{sin}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{2}} {x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{sin}^{\mathrm{6}} {x}\:+\:\mathrm{sin}^{\mathrm{4}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}\:+\:\mathrm{2sin}^{\mathrm{4}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}\:\:+\:\mathrm{2sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{4}} {x}\:+\:\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{4}} {x}\:+\:\mathrm{cos}^{\mathrm{6}} {x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{sin}^{\mathrm{6}} {x}\:+\:\mathrm{cos}^{\mathrm{6}} {x}\:+\:\mathrm{3sin}^{\mathrm{4}} \:{x}\mathrm{cos}^{\mathrm{2}} {x}\:+\:\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{4}} {x} \\ $$$$\Rightarrow\:\mathrm{1}\:−\:\mathrm{3}\:\mathrm{sin}^{\mathrm{4}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}\:−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{4}} {x}\:=\:\mathrm{sin}^{\mathrm{6}} {x}\:+\:\mathrm{cos}^{\mathrm{6}} {x} \\ $$$$\Rightarrow \\ $$$$\:\mathrm{sin}^{\mathrm{6}} \:{x}\:+\:\mathrm{cos}^{\mathrm{6}} \:{x}\:=\:\mathrm{1}−\mathrm{3sin}^{\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\left(\:\mathrm{sin}^{\mathrm{2}} {x}\:+\:\mathrm{cos}^{\mathrm{2}} {x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}−\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x} \\ $$$$\Rightarrow\mathrm{1}\:−\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}\:=\:\frac{\mathrm{7}}{\mathrm{16}}\:;\:{x}\:\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:−\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}\:=\:−\frac{\mathrm{9}}{\mathrm{16}} \\ $$$$\mathrm{or}\:\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{or}\:\:\mathrm{2}{x}\:=\:\pi{n}\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{3}} \\ $$$$\:{x}\:=\:\frac{\pi{n}}{\mathrm{2}}\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{6}} \\ $$$$\:{n}\:=\:\mathrm{0},\:{x}\:=\:\frac{\pi}{\mathrm{6}} \\ $$$${n}\:=\:\mathrm{1},\:{x}\:=\:\frac{\pi}{\mathrm{3}} \\ $$$${n}\:=\:\mathrm{2},\:{x}\:=\:\mathrm{out}\:\mathrm{of}\:\mathrm{range}\:\mathrm{given}\:\Rightarrow\:\mathrm{solution}\:\mathrm{set}\:{S}\:=\:\left\{\frac{\pi}{\mathrm{6}},\frac{\pi}{\mathrm{3}}\right\} \\ $$

Answered by Dwaipayan Shikari last updated on 11/Aug/20

(sin^2 x+cos^2 x)^3 −3sin^2 xcos^2 x(sin^2 x+cos^2 x)=(7/(16)) 1−3cos^2 xsin^2 x=(7/(16)) cos^2 xsin^2 x=(3/(16)) 4cos^2 xsin^2 x=(3/4) sin^2 2x=(3/4) sin2x=±((√3)/2) x=πk±(π/3) x=((πk)/2)±(π/6) (k∈Z) { ((x=(π/6))),((x=(π/3))) :}

$$\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}} −\mathrm{3}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)=\frac{\mathrm{7}}{\mathrm{16}} \\ $$$$\mathrm{1}−\mathrm{3}{cos}^{\mathrm{2}} {xsin}^{\mathrm{2}} {x}=\frac{\mathrm{7}}{\mathrm{16}} \\ $$$${cos}^{\mathrm{2}} {xsin}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} {xsin}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${sin}^{\mathrm{2}} \mathrm{2}{x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${sin}\mathrm{2}{x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${x}=\pi{k}\pm\frac{\pi}{\mathrm{3}} \\ $$$${x}=\frac{\pi{k}}{\mathrm{2}}\pm\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:\:\left({k}\in\mathbb{Z}\right) \\ $$$$\begin{cases}{{x}=\frac{\pi}{\mathrm{6}}}\\{{x}=\frac{\pi}{\mathrm{3}}}\end{cases} \\ $$

Answered by Sarah85 last updated on 11/Aug/20

s^6 +(1−s^2 )^(6/2) =7/16 s^4 −s^2 +3/16=0 sin x =±(√3)/2∨±1/2 x=π/6∨x=π/3

$$\mathrm{s}^{\mathrm{6}} +\left(\mathrm{1}−\mathrm{s}^{\mathrm{2}} \right)^{\mathrm{6}/\mathrm{2}} =\mathrm{7}/\mathrm{16} \\ $$$$\mathrm{s}^{\mathrm{4}} −\mathrm{s}^{\mathrm{2}} +\mathrm{3}/\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:=\pm\sqrt{\mathrm{3}}/\mathrm{2}\vee\pm\mathrm{1}/\mathrm{2} \\ $$$${x}=\pi/\mathrm{6}\vee{x}=\pi/\mathrm{3} \\ $$

Answered by john santu last updated on 11/Aug/20

⋇JS⋇ sin^6 x+cos^6 x=(sin^2 x)^3 +(cos^2 x)^3 =(sin^2 x+cos^2 x)^3 −3sin^2 x cos^2 x(sin^2 x+cos^2 x) =1−(3/4)sin^2 2x (⊸)(7/(16)) = 1−(3/4)sin^2 2x 7 = 16−12sin^2 2x ⇒12sin^2 2x−9=0 ⇒sin 2x = ± (3/(2(√3))) =± ((√3)/2) case(1) ⇒sin 2x = sin 60° ; x=30° 2x=180°−60°+k.360° x = 60°+k.180° ; x=60° case(2)sin 2x=sin (−60°) x=−30°+k.180° ; x = 150° (ignoring) Therefore the solution x=30° ; 60°

$$\:\:\:\:\:\:\divideontimes\mathbb{JS}\divideontimes \\ $$$$\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{cos}\:^{\mathrm{6}} {x}=\left(\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} −\mathrm{3sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}\: \\ $$$$\left(\multimap\right)\frac{\mathrm{7}}{\mathrm{16}}\:=\:\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\mathrm{7}\:=\:\mathrm{16}−\mathrm{12sin}\:^{\mathrm{2}} \mathrm{2}{x}\:\Rightarrow\mathrm{12sin}\:^{\mathrm{2}} \mathrm{2}{x}−\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}\:=\:\pm\:\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\pm\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${case}\left(\mathrm{1}\right)\:\Rightarrow\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{sin}\:\mathrm{60}°\:;\:{x}=\mathrm{30}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{x}=\mathrm{180}°−\mathrm{60}°+{k}.\mathrm{360}°\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:=\:\mathrm{60}°+{k}.\mathrm{180}°\:;\:{x}=\mathrm{60}° \\ $$$${case}\left(\mathrm{2}\right)\mathrm{sin}\:\mathrm{2}{x}=\mathrm{sin}\:\left(−\mathrm{60}°\right) \\ $$$${x}=−\mathrm{30}°+{k}.\mathrm{180}°\:;\:{x}\:=\:\mathrm{150}° \\ $$$$\left({ignoring}\right) \\ $$$${Therefore}\:{the}\:{solution}\: \\ $$$${x}=\mathrm{30}°\:;\:\mathrm{60}°\: \\ $$

Answered by Her_Majesty last updated on 11/Aug/20

sin^6 x+cos^6 x−(7/(16))=0 (3/8)cos4x+(3/(16))=0 cos4x=−(1/2) x=((nπ)/2)±(π/6) in the given interval we get x=(π/6) or x=(π/3)

$${sin}^{\mathrm{6}} {x}+{cos}^{\mathrm{6}} {x}−\frac{\mathrm{7}}{\mathrm{16}}=\mathrm{0} \\ $$$$\frac{\mathrm{3}}{\mathrm{8}}{cos}\mathrm{4}{x}+\frac{\mathrm{3}}{\mathrm{16}}=\mathrm{0} \\ $$$${cos}\mathrm{4}{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\frac{{n}\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{6}} \\ $$$${in}\:{the}\:{given}\:{interval}\:{we}\:{get} \\ $$$${x}=\frac{\pi}{\mathrm{6}}\:{or}\:{x}=\frac{\pi}{\mathrm{3}} \\ $$

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