Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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°

$${an}\:{other}\:{way}: \\ $$$$\left(\mathrm{3sin}\:\theta−\mathrm{4cos}\:\theta\right)\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3sin}\:\theta−\mathrm{4cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta+\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=−\mathrm{1} \\ $$$$\Rightarrow\theta={n}\pi+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}\:{or}\:\mathrm{180}{n}+\mathrm{53}.\mathrm{1}° \\ $$$$\Rightarrow\theta={n}\pi−\frac{\pi}{\mathrm{4}}\:{or}\:\mathrm{180}{n}−\mathrm{45}° \\ $$$$ \\ $$$${within}\:\mathrm{0}−\mathrm{360}°: \\ $$$$\theta=\mathrm{53}.\mathrm{1}°,\:\mathrm{135}°,\:\mathrm{233}.\mathrm{1}°,\:\mathrm{315}° \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com