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Question : 23 of 160
Marks:
+1,
-0
Solution:
We have,
‌‌‌cos‌θ+cos‌2‌θ−√3(sin‌θ+sin‌2θ)+1=0‌⇒‌‌cos‌θ+2cos2θ−√3sin‌θ−2√3sin‌θ‌cos‌θ=0‌⇒‌‌(cos‌θ−√3sin‌θ)+(2cos2θ−2√3sin‌θ‌cos‌θ)=0‌⇒1(cos‌θ−√3sin‌θ)+2‌cos‌θ(cos‌θ−√3sin‌θ)=0‌⇒‌‌(cos‌θ−√3sin‌θ)(2‌cos‌θ+1)=0‌⇒cos‌θ−√3sin‌θ=0‌ or ‌2‌cos‌θ=−1‌tan‌θ=‌‌ or ‌cos‌θ=−‌=cos‌‌‌‌θ=nπ+‌‌‌‌ or ‌θ=2nπ+‌‌‌‌θ=‌,‌‌‌‌ or ‌θ=‌,‌∵ Total number of solution in the interval
(0,2Ï€) is 4 .
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