(1) ((∣a^2 −16∣)/(4−a)) − ((∣a^2 −9∣)/(3+a)) − ((∣4−a^2 ∣)/(2−a)) =? (2)(((∣x∣ +x )/(x−1)))^2 −((14x)/(x−1)) + 12 = 0 (3) log _4 (2^(2x) −(√3) cos x−6sin^2 x ) = x where ((5π)/2) ≤x≤4π (4) ((2cos^2 x−(√3) cos x)/(log


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Question Number 117938 by bemath last updated on 14/Oct/20

$(1) \frac{∣ a^{2} - 16 ∣}{4 - a} - \frac{∣ a^{2} - 9 ∣}{3 + a} - \frac{∣ 4 - a^{2} ∣}{2 - a} = ?$ $(2) {(\frac{∣ x ∣ + x}{x - 1})}^{2} - \frac{14 x}{x - 1} + 12 = 0$ $(3) \log_{4} (2^{2 x} - \sqrt{3} \cos x - 6 \sin^{2} x) = x$ $where \frac{5 π}{2} ⩽ x ⩽ 4 π$ $(4) \frac{2 \cos^{2} x - \sqrt{3} \cos x}{\log_{4} (\sin x)} = 0, where$ $- 3 π ⩽ x ⩽ - \frac{3 π}{2}$
	Answered by john santu last updated on 14/Oct/20

	$(1) \frac{∣ (a - 4) (a + 4) ∣}{4 - a} - \frac{∣ (a + 3) (a - 3) ∣}{3 + a} - \frac{∣ (a - 2) (a + 2) ∣}{2 - a} =$ $f o r a < - 4 \lor a > 4, g i v e s$ $\frac{(a - 4) (a + 4)}{4 - a} - \frac{(a + 3) (a - 3)}{a + 3} - \frac{(a - 2) (a + 2)}{2 - a} =$ $- a - 4 - a + 3 + a + 2 = 1 - a$ $f o r - 2 < a < 2, g i v e s$ $\frac{(4 - a) (4 + a)}{4 - a} - \frac{(3 - a) (3 + a)}{3 + a} - \frac{(2 - a) (2 + a)}{2 - a} =$ $4 + a - 3 + a - 2 - a = - 1 + a$
	Answered by john santu last updated on 14/Oct/20
	$(2)(((∣x∣+x)/(x−1)))^2 −((14x)/(x−1)) +12 = 0 for x < 0 ⇒∣x∣ = −x ⇒−((14x)/(x−1)) = −12 ; 7x = 6x−6 ⇒ x = −6 . for x ≥ 0 ⇒∣x∣ = x ⇒((4x^2 )/((x−1)^2 ))−((14x(x−1))/((x−1)^2 )) +12 = 0; x≠1 2x^2 −7x^2 +7x+6(x^2 −2x+1)=0 x^2 −5x+6 = 0 ; (x−3)(x−2)=0 x = 3 ∧ x=2 . Thus solution set is { −6, 2, 3 }$
	$(2) {(\frac{∣ x ∣ + x}{x - 1})}^{2} - \frac{14 x}{x - 1} + 12 = 0$ $f o r x < 0 \Rightarrow∣ x ∣ = - x$ $\Rightarrow - \frac{14 x}{x - 1} = - 12; 7 x = 6 x - 6$ $\Rightarrow x = - 6 .$ $f o r x ⩾ 0 \Rightarrow∣ x ∣ = x$ $\Rightarrow \frac{4 x^{2}}{{(x - 1)}^{2}} - \frac{14 x (x - 1)}{{(x - 1)}^{2}} + 12 = 0; x \neq 1$ $2 x^{2} - 7 x^{2} + 7 x + 6 (x^{2} - 2 x + 1) = 0$ $x^{2} - 5 x + 6 = 0; (x - 3) (x - 2) = 0$ $x = 3 \land x = 2 . T h u s s o l u t i o n s e t i s$ ${- 6, 2, 3}$
	Answered by john santu last updated on 14/Oct/20

	$(3) \log_{4} (2^{2 x} - \sqrt{3} \cos x - 6 \sin^{2} x) = x$ $\Rightarrow 2^{2 x} - \sqrt{3} \cos x - 6 \sin^{2} x = 4^{x}$ $\Rightarrow - 6 (1 - \cos^{2} x) - \sqrt{3} \cos x = 0$ $\Rightarrow 6 \cos^{2} x - \sqrt{3} \cos x - 6 = 0$ $\Rightarrow (3 \cos x - 2 \sqrt{3}) (2 \cos x + \sqrt{3}) = 0$ $\Rightarrow \cos x = - \frac{\sqrt{3}}{2}$ $\Rightarrow x = \pm \frac{5 π}{6} + 2 k π$
	Answered by Lordose last updated on 14/Oct/20
	$3. 4^x −(√3)cosx−6(1−cos^2 x)=4^x 6cos^2 x−(√3)cosx−6=0 (u−((√3)/(12)))^2 = ((147)/(144)) {u=cosx} u=(((√3)±(√(147)))/(144)) u= −0.8660 or u= 1.1547 ∴ x= cos^(−1) (−0.8660) or cos^(−1) (1.1547) x = 135° or x= complex x=((3π)/4) +2𝛑n {n∈N}$
	$3 .$ $4^{x} - \sqrt{3} cosx - 6 (1 - \cos^{2} x) = 4^{x}$ ${6 \cos}^{2} x - \sqrt{3} cosx - 6 = 0$ ${(u - \frac{\sqrt{3}}{12})}^{2} = \frac{147}{144} {u = cosx}$ $u = \frac{\sqrt{3} \pm \sqrt{147}}{144}$ $u = - 0.8660 or u = 1.1547$ $∴ x = \cos^{- 1} (- 0.8660) or \cos^{- 1} (1.1547)$ $x = 135 ° or x = complex$ $x = \frac{3 π}{4} + 2 π n {n \in N}$
	Commented by bemath last updated on 14/Oct/20

	$\cos^{- 1} (- 0.8660) = 150 ° not 135 ° sir$
	Commented by Lordose last updated on 14/Oct/20

	$Ohh Thanks$
	Answered by john santu last updated on 14/Oct/20
	$(4)2cos^2 x−(√3) cos x = 0 ∧ sin x > 0 cos x(2cos x−(√3)) = 0 ∧ sin x >0 { ((cos x=0)),((cos x=((√3)/2))) :} ∧ −((5π)/2)<x<−((3π)/2)$
	$(4) 2 \cos^{2} x - \sqrt{3} \cos x = 0 \land \sin x > 0$ $\cos x (2 \cos x - \sqrt{3}) = 0 \land \sin x > 0$ ${\begin{cases} \cos x = 0 \\ \cos x = \frac{\sqrt{3}}{2} \end{cases} \land - \frac{5 π}{2} < x < - \frac{3 π}{2}$
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