Q59 / 150 · MHT CET 2024 (16 May Shift 1) · Mathematics

The general solution of the equation $\sqrt{3} \cos \theta+\sin \theta=\sqrt{2}$ is

  • A
    $\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{2}+\frac{\pi}{6}, \mathrm{n} \in \mathbb{Z}$
  • B
    $\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{2}-\frac{\pi}{6}, \mathrm{n} \in \mathbb{Z}$
  • Correct answer
    C
    $\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}-\frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}$
  • D
    $\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}+\frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}$

Solution

$\begin{aligned} & \sqrt{3} \cos \theta+\sin \theta=\sqrt{2} \\ & \Rightarrow \frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\frac{\sqrt{2}}{2} \\ & \Rightarrow \sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta=\frac{1}{\sqrt{2}} \\ & \Rightarrow \sin \left(\theta+\frac{\pi}{3}\right)=\sin \frac{\pi}{4} \\ & \Rightarrow \theta+\frac{\pi}{3}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}, \mathrm{n} \in \mathrm{Z} \\ & \Rightarrow \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}-\frac{\pi}{3}, \mathrm{n} \in \mathrm{Z}\end{aligned}$

Hence, the correct answer is option: 'C'.

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