wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve the following equation:
cos2x+3sinx=2

Open in App
Solution

cos2x+3sinx=2
Put cos2x=12sin2x and we get
2sin2x3sinx+1=0
2sin2x2sinxsinx+1=0
2sinx(sinx1)1(sinx1)=0
(2sinx1)(sinx1)=0
Either, sinx=1,
x=2nπ+π6.
or, sinx=12, x=nπ+(1)nπ6.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Pythagorean Identities
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon