Explanation
We have,
2+3isinθ1−2isinθ
On rationalize and we get,
2+3isinθ1−2isinθ×1+2isinθ1+2isinθ
⇒2+3isinθ+4isinθ+6i2sin2θ12−4i2sin2θ
⇒2−sin2θ+7isinθ1+4sin2θ
⇒1+1−sin2θ+7isinθ1+4sin2θ
⇒1+cos2θ+7isinθ1+4sin2θ
⇒1+cos2θ1+4sin2θ+7isinθ1+4sin2θ
⇒1+cos2θ1+4sin2θ+i7sinθ1+4sin2θ
Now,
Then purely imaginary is
Real part of equal to zero.
1+cos2θ1+4sin2θ=0
⇒1+cos2θ=0
⇒cos2θ=−1
⇒cosθ=√−1
⇒θ=cos−1√−1
Hence, this is the answer.
Step -1: Considering option A.
x2−4x+3√2=0
∴D=(−4)2−4(1)(3√2)
∴D=16−12√2<0
Therefore, it has no real roots.
Step -2: Considering option B.
x2+4x−3√2=0
∴D=(4)2−4(1)(−3√2)
∴D=16+12√2>0
Therefore, it has real roots.
Step -3: Considering option C.
x2−4x−3√2=0
∴D=(−4)2−4(1)(−3√2)
Step -4: Considering option D.
3x2+4√3x+4=0
∴D=(4√3)2−4(3)(4)
∴D=48−48
∴D=0
Therefore, it has real and equal roots.
Hence, option A. is correct answer.
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