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Question Number 162776 by john_santu last updated on 01/Jan/22

((sin x)/(1−cos x)) = (1/2) x=?

$$\:\:\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:{x}=? \\ $$

Answered by bobhans last updated on 01/Jan/22

2sin x + cos x = 1 (√5) ((2/( (√5))) sin x + (1/( (√5))) cos x)= 1 cos (x−α) = (1/( (√5))) ; α=arctan (2) x= ± arccos ((1/( (√5))))+arctan (2)+2kπ ,k∈Z

$$\:\:\:\mathrm{2sin}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{1}\: \\ $$$$\:\:\sqrt{\mathrm{5}}\:\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\mathrm{sin}\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\mathrm{cos}\:\mathrm{x}\right)=\:\mathrm{1} \\ $$$$\:\mathrm{cos}\:\left(\mathrm{x}−\alpha\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:;\:\alpha=\mathrm{arctan}\:\left(\mathrm{2}\right) \\ $$$$\:\:\mathrm{x}=\:\pm\:\mathrm{arccos}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)+\mathrm{arctan}\:\left(\mathrm{2}\right)+\mathrm{2k}\pi\:,\mathrm{k}\in\mathbb{Z} \\ $$

Answered by Rasheed.Sindhi last updated on 01/Jan/22

((sin x)/(1−cos x)) = (1/2) ; x=? 2sin x=1−cos x 4sin^2 x=1+cos^2 x−2cos x 4(1−cos^2 x)=1+cos^2 x−2cos x 5cos^2 x−2cos x−3=0 5cos^2 x−5cos x+3cos x−3=0 5cos x(cos x−1)+3(cos x−1)=0 cos x=1 _(x=2nπ_((Extraneous)) ) ∣ cos x=−(3/5)_(x=cos^(−1) (((−3)/5))^() +2nπ_( _(x≈126.87+2nπ) ) )

$$\:\:\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:;\:{x}=? \\ $$$$\mathrm{2sin}\:{x}=\mathrm{1}−\mathrm{cos}\:{x} \\ $$$$\mathrm{4sin}^{\mathrm{2}} {x}=\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{2cos}\:{x} \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right)=\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{2cos}\:{x} \\ $$$$\mathrm{5cos}^{\mathrm{2}} {x}−\mathrm{2cos}\:{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{5cos}^{\mathrm{2}} {x}−\mathrm{5cos}\:{x}+\mathrm{3cos}\:{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{5cos}\:{x}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)+\mathrm{3}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\underset{\underset{\left(\mathcal{E}{xtraneous}\right)} {{x}=\mathrm{2}{n}\pi}} {\mathrm{cos}\:{x}=\mathrm{1}\:}\mid\underset{\underset{\underset{{x}\approx\mathrm{126}.\mathrm{87}+\mathrm{2}{n}\pi} {\:\:\:\:}} {\overset{} {{x}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{−\mathrm{3}}{\mathrm{5}}\right)}}+\mathrm{2}{n}\pi} {\:\mathrm{cos}\:{x}=−\frac{\mathrm{3}}{\mathrm{5}}} \\ $$

Answered by Berlindo last updated on 02/Jan/22

((2sin (x/2)∙cos (x/2))/(2sin^2 (x/2)))=(1/2) cot (x/2)=1/2⇔tan (x/2)=2 x/2=tan^(−1) 2+πn, n∈Z x=2tan^(−1) 2+2πn, n∈Z

$$\frac{\mathrm{2sin}\:\left(\mathrm{x}/\mathrm{2}\right)\centerdot\mathrm{cos}\:\left(\mathrm{x}/\mathrm{2}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{x}/\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cot}\:\left(\mathrm{x}/\mathrm{2}\right)=\mathrm{1}/\mathrm{2}\Leftrightarrow\mathrm{tan}\:\left(\mathrm{x}/\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{x}/\mathrm{2}=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\pi\mathrm{n},\:\mathrm{n}\in\mathbb{Z} \\ $$$$\mathrm{x}=\mathrm{2tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{2}\pi\mathrm{n},\:\mathrm{n}\in\mathbb{Z} \\ $$

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