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


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

$⊨ B e M a t h ⊨$ $\sin^{6} x + \cos^{6} x = \frac{7}{16}; x \in (0, \frac{π}{2})$
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

$t h a n k y o u a l l s i r$
	Answered by Rio Michael last updated on 11/Aug/20
	$(sin^2 x + cos^2 x)^3 = (sin^2 x + cos^2 x)(sin^2 x + cos^2 x)(sin^2 x + cos^2 x) = (sin^4 x + 2sin^2 x cos^2 x + cos^4 x)(sin^2 x + cos^2 x) = sin^6 x + sin^4 x cos^2 x + 2sin^4 x cos^2 x + 2sin^2 x cos^4 x + sin^2 x cos^4 x + cos^6 x = sin^6 x + cos^6 x + 3sin^4 xcos^2 x + 3 sin^2 x cos^4 x ⇒ 1 − 3 sin^4 x cos^2 x −3 sin^2 x cos^4 x = sin^6 x + cos^6 x ⇒ sin^6 x + cos^6 x = 1−3sin^2 x cos^2 x( sin^2 x + cos^2 x) = 1− (3/4) sin^2 2x ⇒1 − (3/4) sin^2 2x = (7/(16)) ; x ∈(0,(π/2)) −(3/4) sin^2 2x = −(9/(16)) or sin^2 2x = (3/4) ⇒ sin 2x = ((√3)/2) or 2x = πn + (−1)^n (π/3) x = ((πn)/2) + (−1)^n (π/6) n = 0, x = (π/6) n = 1, x = (π/3) n = 2, x = out of range given ⇒ solution set S = {(π/6),(π/3)}$
	${(\sin^{2} x + \cos^{2} x)}^{3} = (\sin^{2} x + \cos^{2} x) (\sin^{2} x + \cos^{2} x) (\sin^{2} x + \cos^{2} x)$ $= (\sin^{4} x + {2 \sin}^{2} x \cos^{2} x + \cos^{4} x) (\sin^{2} x + \cos^{2} x)$ $= \sin^{6} x + \sin^{4} x \cos^{2} x + {2 \sin}^{4} x \cos^{2} x + {2 \sin}^{2} x \cos^{4} x + \sin^{2} x \cos^{4} x + \cos^{6} x$ $= \sin^{6} x + \cos^{6} x + {3 \sin}^{4} x \cos^{2} x + 3 \sin^{2} x \cos^{4} x$ $\Rightarrow 1 - 3 \sin^{4} x \cos^{2} x - 3 \sin^{2} x \cos^{4} x = \sin^{6} x + \cos^{6} x$ $\Rightarrow$ $\sin^{6} x + \cos^{6} x = 1 - {3 \sin}^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)$ $= 1 - \frac{3}{4} \sin^{2} 2 x$ $\Rightarrow 1 - \frac{3}{4} \sin^{2} 2 x = \frac{7}{16}; x \in (0, \frac{π}{2})$ $- \frac{3}{4} \sin^{2} 2 x = - \frac{9}{16}$ $or \sin^{2} 2 x = \frac{3}{4}$ $\Rightarrow \sin 2 x = \frac{\sqrt{3}}{2} or 2 x = π n + {(- 1)}^{n} \frac{π}{3}$ $x = \frac{π n}{2} + {(- 1)}^{n} \frac{π}{6}$ $n = 0, x = \frac{π}{6}$ $n = 1, x = \frac{π}{3}$ $n = 2, x = out of range given \Rightarrow solution set S = {\frac{π}{6}, \frac{π}{3}}$
	Answered by Dwaipayan Shikari last updated on 11/Aug/20
	$(sin^2 x+cos^2 x)^3 −3sin^2 xcos^2 x(sin^2 x+cos^2 x)=(7/(16)) 1−3cos^2 xsin^2 x=(7/(16)) cos^2 xsin^2 x=(3/(16)) 4cos^2 xsin^2 x=(3/4) sin^2 2x=(3/4) sin2x=±((√3)/2) x=πk±(π/3) x=((πk)/2)±(π/6) (k∈Z) { ((x=(π/6))),((x=(π/3))) :}$
	${({s i n}^{2} x + {c o s}^{2} x)}^{3} - 3 {s i n}^{2} {x c o s}^{2} x ({s i n}^{2} x + {c o s}^{2} x) = \frac{7}{16}$ $1 - 3 {c o s}^{2} {x s i n}^{2} x = \frac{7}{16}$ ${c o s}^{2} {x s i n}^{2} x = \frac{3}{16}$ $4 {c o s}^{2} {x s i n}^{2} x = \frac{3}{4}$ ${s i n}^{2} 2 x = \frac{3}{4}$ $s i n 2 x = \pm \frac{\sqrt{3}}{2}$ $x = π k \pm \frac{π}{3}$ $x = \frac{π k}{2} \pm \frac{π}{6} (k \in Z)$ ${\begin{cases} x = \frac{π}{6} \\ x = \frac{π}{3} \end{cases}$
	Answered by Sarah85 last updated on 11/Aug/20

	$s^{6} + {(1 - s^{2})}^{6 / 2} = 7 / 16$ $s^{4} - s^{2} + 3 / 16 = 0$ $\sin x = \pm \sqrt{3} / 2 \lor \pm 1 / 2$ $x = π / 6 \lor x = π / 3$
	Answered by john santu last updated on 11/Aug/20

	$⋇ JS ⋇$ $\sin^{6} x + \cos^{6} x = {(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}$ $= {(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)$ $= 1 - \frac{3}{4} \sin^{2} 2 x$ $(⊸) \frac{7}{16} = 1 - \frac{3}{4} \sin^{2} 2 x$ $7 = 16 - 12 \sin^{2} 2 x \Rightarrow 12 \sin^{2} 2 x - 9 = 0$ $\Rightarrow \sin 2 x = \pm \frac{3}{2 \sqrt{3}} = \pm \frac{\sqrt{3}}{2}$ $c a s e (1) \Rightarrow \sin 2 x = \sin 60 °; x = 30 °$ $2 x = 180 ° - 60 ° + k . 360 °$ $x = 60 ° + k . 180 °; x = 60 °$ $c a s e (2) \sin 2 x = \sin (- 60 °)$ $x = - 30 ° + k . 180 °; x = 150 °$ $(i g n o r i n g)$ $T h e r e f o r e t h e s o l u t i o n$ $x = 30 °; 60 °$
	Answered by Her_Majesty last updated on 11/Aug/20

	${s i n}^{6} x + {c o s}^{6} x - \frac{7}{16} = 0$ $\frac{3}{8} c o s 4 x + \frac{3}{16} = 0$ $c o s 4 x = - \frac{1}{2}$ $x = \frac{n π}{2} \pm \frac{π}{6}$ $i n t h e g i v e n i n t e r v a l w e g e t$ $x = \frac{π}{6} o r x = \frac{π}{3}$
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