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Question Number 103644 by dw last updated on 16/Jul/20

Commented by dw last updated on 16/Jul/20

$[T r i g o n o m e t r i c S u b s t i t .]$
	Answered by 1549442205 last updated on 16/Jul/20
	$putting x=cosϕ(ϕ∈[0,π])we have ((5cosϕ)/(sinϕ))−6cosϕsinϕ=2⇔5cosϕ−6cosϕsin^2 ϕ=2sinϕ Putting tan(ϕ/2)=t we get ((5(1−t^2 ))/(1+t^2 ))−((6(1−t^2 )4t^2 )/((1+t^2 )^3 ))=((4t)/(1+t^2 )) 5(1−t^4 )(1+t^2 )−24(t^2 −t^4 )=4t(t^4 +2t^2 +1) −5t^6 −5t^4 +5t^2 +5−24t^2 +24t^4 =4t^5 +8t^3 +4t ⇔5t^6 +4t^5 −19t^4 +8t^3 +19t^2 +4t−5=0 ⇔5(t^3 −(1/t^3 ))+4(t^2 +(1/t^2 ))−19(t−(1/t))+8=0(1) Putting (t−(1/t))=y⇒t^3 −(1/t^3 )−3(t−(1/t))=y^3 t^2 +(1/t^2 )=y^2 +2.Hence, (1)⇔5y^3 +4y^2 +8−4y+8=0 ⇔5y^3 +4y^2 −4y+16=0 ⇔(y+2)(5y^2 −6y+8)=0 ⇔y+2=0 (because 5y^2 −6y+8=y^2 +(2y−(3/2))^2 +((23)/4)>0) ⇔y=−2⇔t−(1/t)=−2⇔t^2 +2t−1=0 ⇔t=−1±(√2) i)For t=−1+(√2) ⇒t^2 =3−2(√2) we get x=cosϕ=((1−t^2 )/(1+t^2 ))=((2(√2)−2)/(4−2(√2)))=(((√2)−1)/(2−(√2)))=(1/(√2)) ii)For t=−1−(√2) ⇒t^2 =3+2(√2) we get x=cosϕ=((1−t^2 )/(1+t^2 ))=((−2(1+(√2)))/(2(2+(√(2)))))=((−1)/(√2)) Thus ,x∈{−(1/2);((√2)/(√2))} second way: ((5x)/(√(1−x^2 )))+6x(√(1−x^2 ))=2⇔5x−6x(1−x^2 )=2(√(1−x^2 )) ⇔(6x^3 −x)^2 =4(1−x^2 ) ⇔36x^6 −12x^4 +x^2 =4−4x^2 Putting x^2 =y we get 36y^3 −12y^2 +5y−4=0 ⇔(2y−1)(18y^2 −3y+4)=0 ⇔2y−1=0 because 18y^2 −3y+4=17y^2 +(y−(3/2))^2 +(7/4)>0 ⇔y=(1/2)⇔x^2 =(1/2)⇔x=±((√2)/2)$
	$putting x = \cos φ (φ \in [0, π]) we have$ $\frac{5 \cos φ}{\sin φ} - 6 \cos φ \sin φ = 2 \Leftrightarrow 5 \cos φ - 6 \cos φ \sin^{2} φ = 2 \sin φ$ $Putting \tan \frac{φ}{2} = t we get$ $\frac{5 (1 - t^{2})}{1 + t^{2}} - \frac{6 (1 - t^{2}) {4 t}^{2}}{{(1 + t^{2})}^{3}} = \frac{4 t}{1 + t^{2}}$ $5 (1 - t^{4}) (1 + t^{2}) - 24 (t^{2} - t^{4}) = 4 t (t^{4} + {2 t}^{2} + 1)$ $- {5 t}^{6} - {5 t}^{4} + {5 t}^{2} + 5 - {24 t}^{2} + {24 t}^{4} = {4 t}^{5} + {8 t}^{3} + 4 t$ $\Leftrightarrow {5 t}^{6} + {4 t}^{5} - {19 t}^{4} + {8 t}^{3} + {19 t}^{2} + 4 t - 5 = 0$ $\Leftrightarrow 5 (t^{3} - \frac{1}{t^{3}}) + 4 (t^{2} + \frac{1}{t^{2}}) - 19 (t - \frac{1}{t}) + 8 = 0 (1)$ $Putting (t - \frac{1}{t}) = y \Rightarrow t^{3} - \frac{1}{t^{3}} - 3 (t - \frac{1}{t}) = y^{3}$ $t^{2} + \frac{1}{t^{2}} = y^{2} + 2 . Hence,$ $(1) \Leftrightarrow {5 y}^{3} + {4 y}^{2} + 8 - 4 y + 8 = 0$ $\Leftrightarrow {5 y}^{3} + {4 y}^{2} - 4 y + 16 = 0$ $\Leftrightarrow (y + 2) ({5 y}^{2} - 6 y + 8) = 0$ $\Leftrightarrow y + 2 = 0 (because {5 y}^{2} - 6 y + 8 = y^{2} + {(2 y - \frac{3}{2})}^{2} + \frac{23}{4} > 0)$ $\Leftrightarrow y = - 2 \Leftrightarrow t - \frac{1}{t} = - 2 \Leftrightarrow t^{2} + 2 t - 1 = 0$ $\Leftrightarrow t = - 1 \pm \sqrt{2}$ $i) For t = - 1 + \sqrt{2} \Rightarrow t^{2} = 3 - 2 \sqrt{2} we get$ $x = \cos φ = \frac{1 - t^{2}}{1 + t^{2}} = \frac{2 \sqrt{2} - 2}{4 - 2 \sqrt{2}} = \frac{\sqrt{2} - 1}{2 - \sqrt{2}} = \frac{1}{\sqrt{2}}$ $ii) For t = - 1 - \sqrt{2} \Rightarrow t^{2} = 3 + 2 \sqrt{2} we get$ $x = \cos φ = \frac{1 - t^{2}}{1 + t^{2}} = \frac{- 2 (1 + \sqrt{2})}{2 (2 + \sqrt{2)}} = \frac{- 1}{\sqrt{2}}$ $Thus, x \in {- \frac{1}{2}; \frac{\sqrt{2}}{\sqrt{2}}}$ $second way :$ $\frac{5 x}{\sqrt{1 - x^{2}}} + 6 x \sqrt{1 - x^{2}} = 2 \Leftrightarrow 5 x - 6 x (1 - x^{2}) = 2 \sqrt{1 - x^{2}}$ $\Leftrightarrow {({6 x}^{3} - x)}^{2} = 4 (1 - x^{2})$ $\Leftrightarrow {36 x}^{6} - {12 x}^{4} + x^{2} = 4 - {4 x}^{2}$ $Putting x^{2} = y we get$ ${36 y}^{3} - {12 y}^{2} + 5 y - 4 = 0$ $\Leftrightarrow (2 y - 1) ({18 y}^{2} - 3 y + 4) = 0$ $\Leftrightarrow 2 y - 1 = 0 because {18 y}^{2} - 3 y + 4 = {17 y}^{2} + {(y - \frac{3}{2})}^{2} + \frac{7}{4} > 0$ $\Leftrightarrow y = \frac{1}{2} \Leftrightarrow x^{2} = \frac{1}{2} \Leftrightarrow x = \pm \frac{\sqrt{2}}{2}$
	Commented by maths mind last updated on 16/Jul/20

	$5 - 6 {s i n}^{2} (x) = 2 t g (x) . . .$ $v e r r y n i c e w e c a n u s e m a y b e e t h i s$ $\Rightarrow 5 - \frac{6 {t g}^{2} (x)}{1 + {t g}^{2} (x)} = 2 t g (x)$ $t = t g (x)$ $5 - 6 \frac{t^{2}}{1 + t^{2}} = 2 t$ $5 - t^{2} - 2 t^{3} - 2 t = 0$ $t = 1 e v i d e n t e s o l u t i o n$ $\Leftrightarrow (t - 1) (- 2 t^{2} - 3 t - 5) = 0$
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