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

Solve the following equations.
Solve the equation 4cosx(23sin2x)+(cos2x+1)=0. Find the least distance between its positive roots.

Open in App
Solution

4cosx(23sin2x)+cos2x+1=0
4cosx(23(1cos2x))+2cos2x=0
4cosx(3cos2x1)+2cos2x=0
12cos3x4cosx+2cos2x=0
2cosx(6cos2x+cosx2)=0
2cosx=0
6cos2x+cosx2=0
cox=0
cox=1±1+4812
x=(2x+1)π/2 =1±712
=π/2,3π/2...=1/22/3
x=2xπ±π/3,2xπ±cos1(2/3)
=π/3,....
Least distance b/w +ve roots =π/2π/3
=π/6

1124975_888386_ans_a0c9359cbdfa4580b736330365cc80e1.jpeg

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