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Question : 118 of 150
Marks:
+1,
-0
Solution:
We have,
sin‌x−3‌sin‌2‌x+sin‌3‌x=cos‌x −3‌cos‌2‌x+cos‌3‌x ⇒sin‌x+sin‌3‌x−3‌sin‌2‌x=cos‌x +cos‌3‌x−3‌cos‌2‌x ⇒2‌sin‌2‌x‌cos‌x−3‌sin‌2‌x−2‌cos‌2‌x cos‌x+3‌cos‌2‌x=0 ⇒sin‌2‌x(2‌cos‌x−3)−cos‌2‌x (2‌cos‌x−3)=0 ⇒(sin‌2‌x−cos‌2‌x)(2‌cos‌x−3)=0 ⇒sin‌2‌x=cos‌2‌x[∵cos‌x≠3∕2] ⇒2x=2xπ±(−2x) ⇒x=+[∵ neglect
− ve sign
]
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