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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
The general solution of the equation $7\cos^2\text{x}+3\sin^2\text{x}=4$ is:
  • A
    $\text{x}=2\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
  • B
    $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{Z}$
  • ✓
    $\text{x}=\text{n}\pi\frac{\pi}{3},\text{n}\in\text{Z}$
  • D
    None of these
✓

Answer

Correct option: C.
$\text{x}=\text{n}\pi\frac{\pi}{3},\text{n}\in\text{Z}$
Given:
$7\cos^2\text{x}+3\sin^2\text{x}=4$
$\Rightarrow7\cos^2\text{x}+3\Big(1-\cos^2\text{x}\Big)=4$
$\Rightarrow7\cos^2\text{x}+3-3\cos^2\text{x}=4$
$\Rightarrow4\cos^2\text{x}+3=4$
$\Rightarrow4\Big(1-\cos^2\text{x}\Big)=3$
$\Rightarrow4\sin^2\text{x}=3$
$\Rightarrow\sin^2\text{x}=\frac{3}{4}$
$\Rightarrow\sin\text{x}=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{3}$
$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$

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