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

(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 _4 (sin x))) = 0 , where −3π ≤x≤−((3π)/2)

$$\left(\mathrm{1}\right)\:\frac{\mid{a}^{\mathrm{2}} −\mathrm{16}\mid}{\mathrm{4}−{a}}\:−\:\frac{\mid{a}^{\mathrm{2}} −\mathrm{9}\mid}{\mathrm{3}+{a}}\:−\:\frac{\mid\mathrm{4}−{a}^{\mathrm{2}} \mid}{\mathrm{2}−{a}}\:=? \\ $$$$\left(\mathrm{2}\right)\left(\frac{\mid\mathrm{x}\mid\:+\mathrm{x}\:}{\mathrm{x}−\mathrm{1}}\right)^{\mathrm{2}} −\frac{\mathrm{14x}}{\mathrm{x}−\mathrm{1}}\:+\:\mathrm{12}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{2}^{\mathrm{2x}} −\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}−\mathrm{6sin}\:^{\mathrm{2}} \mathrm{x}\:\right)\:=\:\mathrm{x} \\ $$$$\mathrm{where}\:\frac{\mathrm{5}\pi}{\mathrm{2}}\:\leqslant\mathrm{x}\leqslant\mathrm{4}\pi \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}−\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}}{\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{sin}\:\mathrm{x}\right)}\:=\:\mathrm{0}\:,\:\mathrm{where}\: \\ $$$$−\mathrm{3}\pi\:\leqslant\mathrm{x}\leqslant−\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$

Answered by john santu last updated on 14/Oct/20

(1) ((∣(a−4)(a+4)∣)/(4−a)) −((∣(a+3)(a−3)∣)/(3+a)) −((∣(a−2)(a+2)∣)/(2−a)) = for a<−4 ∨ a > 4 , gives (((a−4)(a+4))/(4−a)) − (((a+3)(a−3))/(a+3)) −(((a−2)(a+2))/(2−a)) = −a−4−a+3+a+2 = 1−a for −2<a<2 , gives (((4−a)(4+a))/(4−a))−(((3−a)(3+a))/(3+a))−(((2−a)(2+a))/(2−a)) = 4+a−3+a−2−a = −1+ a

$$\left(\mathrm{1}\right)\:\frac{\mid\left({a}−\mathrm{4}\right)\left({a}+\mathrm{4}\right)\mid}{\mathrm{4}−{a}}\:−\frac{\mid\left({a}+\mathrm{3}\right)\left({a}−\mathrm{3}\right)\mid}{\mathrm{3}+{a}}\:−\frac{\mid\left({a}−\mathrm{2}\right)\left({a}+\mathrm{2}\right)\mid}{\mathrm{2}−{a}}\:= \\ $$$${for}\:{a}<−\mathrm{4}\:\vee\:{a}\:>\:\mathrm{4}\:,\:{gives}\: \\ $$$$\frac{\left({a}−\mathrm{4}\right)\left({a}+\mathrm{4}\right)}{\mathrm{4}−{a}}\:−\:\frac{\left({a}+\mathrm{3}\right)\left({a}−\mathrm{3}\right)}{{a}+\mathrm{3}}\:−\frac{\left({a}−\mathrm{2}\right)\left({a}+\mathrm{2}\right)}{\mathrm{2}−{a}}\:= \\ $$$$−{a}−\mathrm{4}−{a}+\mathrm{3}+{a}+\mathrm{2}\:=\:\mathrm{1}−{a} \\ $$$${for}\:−\mathrm{2}<{a}<\mathrm{2}\:,\:{gives}\: \\ $$$$\frac{\left(\mathrm{4}−{a}\right)\left(\mathrm{4}+{a}\right)}{\mathrm{4}−{a}}−\frac{\left(\mathrm{3}−{a}\right)\left(\mathrm{3}+{a}\right)}{\mathrm{3}+{a}}−\frac{\left(\mathrm{2}−{a}\right)\left(\mathrm{2}+{a}\right)}{\mathrm{2}−{a}}\:= \\ $$$$\mathrm{4}+{a}−\mathrm{3}+{a}−\mathrm{2}−{a}\:=\:−\mathrm{1}+\:{a} \\ $$$$ \\ $$

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 }

$$\left(\mathrm{2}\right)\left(\frac{\mid{x}\mid+{x}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} −\frac{\mathrm{14}{x}}{{x}−\mathrm{1}}\:+\mathrm{12}\:=\:\mathrm{0} \\ $$$${for}\:{x}\:<\:\mathrm{0}\:\Rightarrow\mid{x}\mid\:=\:−{x}\: \\ $$$$\Rightarrow−\frac{\mathrm{14}{x}}{{x}−\mathrm{1}}\:=\:−\mathrm{12}\:;\:\mathrm{7}{x}\:=\:\mathrm{6}{x}−\mathrm{6} \\ $$$$\Rightarrow\:{x}\:=\:−\mathrm{6}\:. \\ $$$${for}\:{x}\:\geqslant\:\mathrm{0}\:\Rightarrow\mid{x}\mid\:=\:{x} \\ $$$$\Rightarrow\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{14}{x}\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\mathrm{12}\:=\:\mathrm{0};\:{x}\neq\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{6}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}\:=\:\mathrm{0}\:;\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$${x}\:=\:\mathrm{3}\:\wedge\:{x}=\mathrm{2}\:.\:{Thus}\:{solution}\:{set}\:{is} \\ $$$$\left\{\:−\mathrm{6},\:\mathrm{2},\:\mathrm{3}\:\right\} \\ $$$$ \\ $$

Answered by john santu last updated on 14/Oct/20

(3)log _4 (2^(2x) −(√3) cos x−6sin^2 x) = x ⇒ 2^(2x) −(√3) cos x−6sin^2 x = 4^x ⇒−6(1−cos^2 x)−(√3) cos x = 0 ⇒6 cos^2 x−(√(3 )) cos x−6 =0 ⇒(3cos x−2(√3))(2cos x+(√3))=0 ⇒cos x= −((√3)/2) ⇒x = ± ((5π)/6) + 2kπ

$$\left(\mathrm{3}\right)\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{2}^{\mathrm{2}{x}} −\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{6sin}\:^{\mathrm{2}} {x}\right)\:=\:{x} \\ $$$$\Rightarrow\:\mathrm{2}^{\mathrm{2}{x}} −\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{6sin}\:^{\mathrm{2}} {x}\:=\:\mathrm{4}^{{x}} \\ $$$$\Rightarrow−\mathrm{6}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)−\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{6}\:\mathrm{cos}\:^{\mathrm{2}} {x}−\sqrt{\mathrm{3}\:}\:\mathrm{cos}\:{x}−\mathrm{6}\:=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3cos}\:{x}−\mathrm{2}\sqrt{\mathrm{3}}\right)\left(\mathrm{2cos}\:{x}+\sqrt{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}\:=\:\pm\:\frac{\mathrm{5}\pi}{\mathrm{6}}\:+\:\mathrm{2}{k}\pi\: \\ $$$$ \\ $$

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}

$$ \\ $$$$ \\ $$$$\mathrm{3}. \\ $$$$\mathrm{4}^{\mathrm{x}} −\sqrt{\mathrm{3}}\mathrm{cosx}−\mathrm{6}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)=\mathrm{4}^{\mathrm{x}} \\ $$$$\mathrm{6cos}^{\mathrm{2}} \mathrm{x}−\sqrt{\mathrm{3}}\mathrm{cosx}−\mathrm{6}=\mathrm{0} \\ $$$$\left(\mathrm{u}−\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\right)^{\mathrm{2}} =\:\frac{\mathrm{147}}{\mathrm{144}}\:\:\left\{\mathrm{u}=\mathrm{cosx}\right\} \\ $$$$\mathrm{u}=\frac{\sqrt{\mathrm{3}}\pm\sqrt{\mathrm{147}}}{\mathrm{144}} \\ $$$$\mathrm{u}=\:−\mathrm{0}.\mathrm{8660}\:\mathrm{or}\:\mathrm{u}=\:\mathrm{1}.\mathrm{1547} \\ $$$$\therefore\:\mathrm{x}=\:\mathrm{cos}^{−\mathrm{1}} \left(−\mathrm{0}.\mathrm{8660}\right)\:\mathrm{or}\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{1}.\mathrm{1547}\right) \\ $$$$\:\mathrm{x}\:=\:\mathrm{135}°\:\mathrm{or}\:\mathrm{x}=\:\mathrm{complex} \\ $$$$\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{4}}\:+\mathrm{2}\boldsymbol{\pi}\mathrm{n}\:\left\{\mathrm{n}\in\mathbb{N}\right\} \\ $$$$ \\ $$

Commented by bemath last updated on 14/Oct/20

cos^(−1) (−0.8660) = 150° not 135° sir

$$\mathrm{cos}^{−\mathrm{1}} \left(−\mathrm{0}.\mathrm{8660}\right)\:=\:\mathrm{150}°\:\mathrm{not}\:\mathrm{135}°\:\mathrm{sir} \\ $$

Commented by Lordose last updated on 14/Oct/20

Ohh Thanks

$$\mathrm{Ohh}\:\mathrm{Thanks} \\ $$

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)

$$\left(\mathrm{4}\right)\mathrm{2cos}\:^{\mathrm{2}} {x}−\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}\:=\:\mathrm{0}\:\wedge\:\mathrm{sin}\:{x}\:>\:\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\left(\mathrm{2cos}\:{x}−\sqrt{\mathrm{3}}\right)\:=\:\mathrm{0}\:\wedge\:\mathrm{sin}\:{x}\:>\mathrm{0} \\ $$$$\begin{cases}{\mathrm{cos}\:{x}=\mathrm{0}}\\{\mathrm{cos}\:{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases}\:\:\:\:\wedge\:−\frac{\mathrm{5}\pi}{\mathrm{2}}<{x}<−\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$

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