The solution set of the equation 4sinθ−2cosθ−2√3sinθ+√3=0 in the interval (0,2π) is-
Explanation
Simplifying sin3x=cos2x.
sin3x=cos2x
3sinx−4sin3x=1−2sin2x
4sin3x−2sin2x−3sinx+1=0
Put sinx=t.
4t3−2t2−3t+1=0
(t−1)(4t2+2t−1)=0
t=1
Or,
4t2+2t−1=0
t=−2±√(2)2−4(4)(−1)2×4
=−2±√208
=−2±2√58
=−1±√54
Then,
sinx=1
x=π2
sinx=−1±√54
sinx=−1+√54
x=π+π10
x=11π10
sinx=−1−√54
x=2π−π10
x=19π10
So, x=π2, x=11π10 and x=19π10.
Since the given interval is (π2,π), then, x=π2 is the only solution.
Therefore, the number of solutions of sin3x=cos2x is 1.
Please disable the adBlock and continue. Thank you.