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Question Number 180826 by mnjuly1970 last updated on 17/Nov/22
      If  ,   2sin(θ )−3cos(θ) =3   ⇒   2sin((θ/2)) − 3cos((θ/2)) = ?
$$ \\ $$$$\:\:\:\:\mathrm{I}{f}\:\:,\:\:\:\mathrm{2}{sin}\left(\theta\:\right)−\mathrm{3}{cos}\left(\theta\right)\:=\mathrm{3} \\ $$$$\:\Rightarrow\:\:\:\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\:−\:\mathrm{3}{cos}\left(\frac{\theta}{\mathrm{2}}\right)\:=\:? \\ $$$$ \\ $$
Answered by mr W last updated on 17/Nov/22
2 sin θ=3(1+cos θ)  2 sin (θ/2) cos (θ/2)=3 cos^2  (θ/2)  (2 sin (θ/2)−3 cos (θ/2))cos (θ/2)=0  ⇒2 sin (θ/2)−3 cos (θ/2)=0 ✓  or  cos (θ/2)=0 ⇒sin (θ/2)=±1  2 sin (θ/2)−3 cos (θ/2)=±2−0=±2 ✓
$$\mathrm{2}\:\mathrm{sin}\:\theta=\mathrm{3}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$$\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}−\mathrm{3}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}−\mathrm{3}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\mathrm{0}\:\checkmark \\ $$$${or} \\ $$$$\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\pm\mathrm{1} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}−\mathrm{3}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\pm\mathrm{2}−\mathrm{0}=\pm\mathrm{2}\:\checkmark \\ $$

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