Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The solution of $\sin x=-\frac{\sqrt{3}}{2}$ is
MathematicsTrigonometric EquationsVITEEEVITEEE 2019
Options:
  • A $x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{4 \pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$
  • B $x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{2 \pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$
  • C $x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{3 \pi}{3}$, where $\mathrm{n} \in Z$
  • D None of the these
Solution:
2360 Upvotes Verified Answer
The correct answer is: $x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{4 \pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$
We have, $\sin x=-\frac{\sqrt{3}}{2}=-\sin \frac{\pi}{3}$
Hence, $\sin x=\sin \frac{4 \pi}{3}$, which gives $\mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{4 \pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.