If 4sin
2
θ−4
3
sinθ+3=0, then find the value of θ (0
∘
<θ<90
∘
).
- A30
∘ - B75
∘ - C60
∘ - D45
∘
Solution & Step-by-step Explanation
Let sinθ=t. The given equation can be written as a quadratic equation in terms of t:
4t
2
−4
3
t+3=0
Notice that this expression is a perfect square:
(2t−
3
)
2
=0
Taking the square root on both sides:
2t−
3
=0
2t=
3
t=
2
3
Substitute back t=sinθ:
sinθ=
2
3
Since 0
∘
<θ<90
∘
, we know that sin60
∘
=
2
3
.
Therefore, θ=60
∘
.
4t
2
−4
3
t+3=0
Notice that this expression is a perfect square:
(2t−
3
)
2
=0
Taking the square root on both sides:
2t−
3
=0
2t=
3
t=
2
3
Substitute back t=sinθ:
sinθ=
2
3
Since 0
∘
<θ<90
∘
, we know that sin60
∘
=
2
3
.
Therefore, θ=60
∘
.