cos x (√(tan x)) = sin^3 x+cos^3 x


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 136008 by liberty last updated on 17/Mar/21

$\cos x \sqrt{\tan x} = \sin^{3} x + \cos^{3} x$
	Answered by MJS_new last updated on 17/Mar/21

	$t = \sqrt{\tan x} \land t ⩾ 0$ $\frac{t}{{(t^{4} + 1)}^{1 / 2}} = \frac{t^{6} + 1}{{(t^{4} + 1)}^{3 / 2}}$ $t^{6} - t^{5} - t + 1 = 0$ $(t - 1) (t^{5} - 1) = 0 \Rightarrow t = 1$ $\Rightarrow x = \frac{π}{4} + n π$
	Answered by liberty last updated on 18/Mar/21
	$note that tan x ≥ 0 ⇒ (√(sin x cos x)) = (sin x+cos x)(1−sin x cos x) ⇒(√(sin x cos x)) =(√(1+2sin x cos x)) (1−sin x cos x ) let sin x cos x = ℓ ⇒(√ℓ) = (√(1+2ℓ)) (1−ℓ) ⇒ℓ = (1+2ℓ)(1−2ℓ+ℓ^2 ) ⇒ℓ = 1−2ℓ+ℓ^2 +2ℓ−4ℓ^2 +2ℓ^3 ⇒2ℓ^3 −3ℓ^2 −ℓ+1=0 ⇒(2ℓ−1)(ℓ^2 −ℓ−1)=0 ⇒ { ((ℓ=(1/2)⇒sin xcos x=(1/2), sin 2x=1)),((ℓ = ((1±(√5))/2) (rejected))) :} so solution is 2x = (π/2)+2nπ or x = (π/4)+nπ$
	$n o t e t h a t \tan x ⩾ 0$ $\Rightarrow \sqrt{\sin x \cos x} = (\sin x + \cos x) (1 - \sin x \cos x)$ $\Rightarrow \sqrt{\sin x \cos x} = \sqrt{1 + 2 \sin x \cos x} (1 - \sin x \cos x)$ $l e t \sin x \cos x = ℓ$ $\Rightarrow \sqrt{ℓ} = \sqrt{1 + 2 ℓ} (1 - ℓ)$ $\Rightarrow ℓ = (1 + 2 ℓ) (1 - 2 ℓ + ℓ^{2})$ $\Rightarrow ℓ = 1 - 2 ℓ + ℓ^{2} + 2 ℓ - 4 ℓ^{2} + 2 ℓ^{3}$ $\Rightarrow 2 ℓ^{3} - 3 ℓ^{2} - ℓ + 1 = 0$ $\Rightarrow (2 ℓ - 1) (ℓ^{2} - ℓ - 1) = 0$ $\Rightarrow {\begin{cases} ℓ = \frac{1}{2} \Rightarrow \sin x \cos x = \frac{1}{2}, \sin 2 x = 1 \\ ℓ = \frac{1 \pm \sqrt{5}}{2} (r e j e c t e d) \end{cases}$ $s o s o l u t i o n i s 2 x = \frac{π}{2} + 2 n π$ $o r x = \frac{π}{4} + n π$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum