Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Dear student,
As
sinxsin3x=1/2
1/2*(2sinxsin3x)=1/2 ; multipying and dividing by 2,
Therefore 2sinxsin3x=1
using formula 2sinAsinB=cos(A-B)-cos(A+B)
here 2sixsin3x=cos2x-cos4x
cos2x-cos4x=1
so cos2x=1+cos4x
cos2x=2cos22x ; using 1+cos2A=2cos2A
Therefore, 1=2cos2x
so, 1/2=cos2x
so, x=30 as cos60=1/2
All the best!
Please approve my answer if you like it!
sinxsin3x = 1/2
2sinxsin3x = 1
cos-2x - cos4x = 1 (2sinasinb = cosa-b - cosa+b)
cos2x - cos4x = 1 ...........1
cos4x = 2cos22x - 1 so eq 1 becomes
cos2x - (2cos22x-1) = 1
cos2x - 2cos22x = 0
cos2x(1-2cos2x) = 0
cos2x = 0 or 1-2cos2x = 0
2x = (2n+1)pi/2 2x = 2npi +(-)pi/3
x = (2n+1)pi/4 or x = npi +(-)pi/6
these are the required solutions of this equation
approve if u like my ans
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !