CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Number of solutions of the equation 2sinx23cosx3tanx+3=0 where x[0,2π) is


Solution

2sinx23cosx3tanx+3=0
2[sinx3cosx]3cosx[sinx3cosx]=0
[sinx3cosx][23cosx]=0
For sinx3cosx=0
tanx=3
x=π3,4π3
For 23cosx=0
cosx=32
x=π6,11π6
So, x=π6,π3,4π3,11π6

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image