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

Number of solutions of the equation 2(sin3θ+sin2θ)+2(cos3θ+cos2θ) = 3sin2θ
in the interval [0, 4π] is

Open in App
Solution

2(sin3θ+sin2θ)+2(cos3θ+cos2θ) = 3sin2θsin3θ+cos3θ+1 = 3sinθcosθsinθ+cosθ+1 = 0 (a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca))

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