Find the General Solution of the Following Equation: Sec X = √ 2

प्रश्न

Find the general solution of the following equation:

\[\sec x = \sqrt{2}\]

योग

उत्तर

We have:
\[\sec x = \sqrt{2}\] (or)

\[\cos x = \frac{1}{\sqrt{2}}\]

The value of x satisfying \[\cos x = \frac{1}{\sqrt{2}}\] is \[\frac{\pi}{4}\]

∴ \[\cos x = \frac{1}{\sqrt{2}}\]

⇒ \[\cos x = \cos \frac{\pi}{4}\]

⇒ \[x = 2n\pi \pm \frac{\pi}{4}\],

\[n \in Z\]

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Trigonometric Equations

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Q 1.4 Q 1.3 Q 1.5

अध्याय 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २१]

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आर.डी. शर्मा Mathematics [English] Class 11

अध्याय 11 Trigonometric equations
Exercise 11.1 | Q 1.4 | पृष्ठ २१

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Trigonometric Equations

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