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Question : 4 of 90
Marks:
+1,
-0
Solution:
Given,
3(sin‌3‌θ)x−y+z=23(cos‌2‌θ)x+4y+3z=36x+7y+7z=9For no solutions determinant of coefficient will be
=0∴D=|| 3‌sin‌3‌θ | −1 | 1 |
| 3‌cos‌2‌θ | 4 | 3 |
| 6 | 7 | 7 |
|=0⇒3‌sin‌3‌θ(28−21)+1(21‌cos‌2‌θ−18)+1(21‌cos‌2‌θ−24)=0⇒21‌sin‌3‌θ+42‌cos‌2‌θ−42=0⇒sin‌3‌θ+2‌cos‌2‌θ−2=0⇒3‌sin‌θ−4sin3θ+2(1−2sin2θ)−2=0 ⇒3‌sin‌θ−4sin3θ−4sin2θ=0⇒4sin3θ+4sin2θ−3‌sin‌θ=0⇒sin‌θ(4sin2θ+4‌sin‌θ−3)=0∴sin‌θ=0⇒θ=π,2π,3π when
θ∈(0,4π)or,
4‌sin‌‌2‌θ+4‌sin‌θ−3=0⇒4‌sin‌2‌θ+6‌sin‌θ−2‌sin‌θ−3=0⇒2‌sin‌θ(2‌sin‌θ+3)−1(2‌sin‌θ+3)=0⇒(2‌sin‌θ−1)(2‌sin‌θ+3)=0∴sin‌θ=‌ or,
sin‌θ=−‌ [not possible as
sin∈[−1,1] ]
∴sin‌θ=‌⇒θ=‌,‌,‌,‌∴ Possible values of
θ=π,2π,3π,‌,‌,‌,‌∴ Total 7 values of
θ possible.
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