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3sin-2-sin-cos-4cos-2-0-find-the-values-of-if-lies-between-0-and-360-

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Question Number 29354 by NECx last updated on 08/Feb/18
3sin^2 θ −sin θcos θ −4cos^2 θ=0 find the values of θ if θ lies between 0 and 360
$$\mathrm{3sin}\:^{\mathrm{2}} \theta\:−\mathrm{sin}\:\theta\mathrm{cos}\:\theta\:−\mathrm{4cos}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$${find}\:{the}\:{values}\:{of}\:\theta\:{if}\:\theta\:{lies} \\ $$$${between}\:\mathrm{0}\:{and}\:\mathrm{360} \\ $$
Answered by mrW2 last updated on 08/Feb/18
3(1−cos^2 θ) −sin θcos θ −4cos^2 θ=0 3−sin θcos θ −7cos^2 θ=0 6−2sin θcos θ −7(2cos^2 θ)=0 6−sin 2θ −7(1+cos 2θ)=0 sin 2θ+7cos 2θ=−1 sin 2θ×(1/( (√(7^2 +1^1 ))))+cos 2θ×(7/( (√(7^2 +1^2 ))))=−(1/( (√(7^2 +1^2 )))) sin 2θ×sin α+cos 2θ×cos α=−(1/(5(√2))) with α=tan^(−1) (1/7) ⇒cos (2θ−α)=−(1/(5(√2))) ⇒2θ−α=(2n+1)π±cos^(−1) (1/(5(√2))) ⇒θ=nπ+(1/2)(π+α±cos^(−1) (1/(5(√2)))) ⇒θ=nπ+(1/2)(π+tan^(−1) (1/7)±cos^(−1) (1/(5(√2)))) ⇒θ=180n+ { ((135°)),((53.1°)) :} solutions within 0°−360°: ⇒θ=53.1°, 135°, 233.1°, 315°
$$\mathrm{3}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\theta\right)\:−\mathrm{sin}\:\theta\mathrm{cos}\:\theta\:−\mathrm{4cos}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{3}−\mathrm{sin}\:\theta\mathrm{cos}\:\theta\:−\mathrm{7cos}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{6}−\mathrm{2sin}\:\theta\mathrm{cos}\:\theta\:−\mathrm{7}\left(\mathrm{2cos}\:^{\mathrm{2}} \theta\right)=\mathrm{0} \\ $$$$\mathrm{6}−\mathrm{sin}\:\mathrm{2}\theta\:−\mathrm{7}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta+\mathrm{7cos}\:\mathrm{2}\theta=−\mathrm{1} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta×\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{1}^{\mathrm{1}} }}+\mathrm{cos}\:\mathrm{2}\theta×\frac{\mathrm{7}}{\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta×\mathrm{sin}\:\alpha+\mathrm{cos}\:\mathrm{2}\theta×\mathrm{cos}\:\alpha=−\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{cos}\:\left(\mathrm{2}\theta−\alpha\right)=−\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{2}\theta−\alpha=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\theta={n}\pi+\frac{\mathrm{1}}{\mathrm{2}}\left(\pi+\alpha\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{2}}}\right) \\ $$$$\Rightarrow\theta={n}\pi+\frac{\mathrm{1}}{\mathrm{2}}\left(\pi+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{7}}\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{2}}}\right) \\ $$$$\Rightarrow\theta=\mathrm{180}{n}+\begin{cases}{\mathrm{135}°}\\{\mathrm{53}.\mathrm{1}°}\end{cases} \\ $$$${solutions}\:{within}\:\mathrm{0}°−\mathrm{360}°: \\ $$$$\Rightarrow\theta=\mathrm{53}.\mathrm{1}°,\:\mathrm{135}°,\:\mathrm{233}.\mathrm{1}°,\:\mathrm{315}° \\ $$
Answered by mrW2 last updated on 08/Feb/18
an other way: (3sin θ−4cos θ)(sin θ+cos θ)=0 ⇒3sin θ−4cos θ=0 ⇒sin θ+cos θ=0 ⇒tan θ=(4/3) ⇒tan θ=−1 ⇒θ=nπ+tan^(−1) (4/3) or 180n+53.1° ⇒θ=nπ−(π/4) or 180n−45° within 0−360°: θ=53.1°, 135°, 233.1°, 315°
$${an}\:{other}\:{way}: \\ $$$$\left(\mathrm{3sin}\:\theta−\mathrm{4cos}\:\theta\right)\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3sin}\:\theta−\mathrm{4cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta+\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=−\mathrm{1} \\ $$$$\Rightarrow\theta={n}\pi+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}\:{or}\:\mathrm{180}{n}+\mathrm{53}.\mathrm{1}° \\ $$$$\Rightarrow\theta={n}\pi−\frac{\pi}{\mathrm{4}}\:{or}\:\mathrm{180}{n}−\mathrm{45}° \\ $$$$ \\ $$$${within}\:\mathrm{0}−\mathrm{360}°: \\ $$$$\theta=\mathrm{53}.\mathrm{1}°,\:\mathrm{135}°,\:\mathrm{233}.\mathrm{1}°,\:\mathrm{315}° \\ $$

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