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Question 1 of 3
Given that 0≤θ≤2π, solve for θ
tan2θ=3
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1.
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2.
π6,5π6,7π6,11π6
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3.
π3,4π3
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4.
π3,2π3,4π3,5π3
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First, simplify the given equation
tan2θ |
= |
3 |
√tan2θ |
= |
√3 |
Get the square root of both sides |
tanθ |
= |
±√3 |
Remember that tanθ=sinθcosθ. Knowing that tanθ can be either positive or negative, this means that θ lies on all 4 quadrants
Next, find the reference angle to work with by using special triangles
tanθ |
= |
±√3 |
= |
oppositeadjacent |
|
tanθ |
= |
±√3 |
= |
√31 |
Using π3 as θ |
|
tanθ |
= |
±√3 |
= |
√3 |
Hence, the reference angle and the value of θ on the First Quadrant is π3
Finally, use the reference angle to find the value of θ on the other 3 quadrants
Remember that the reference angle is always relative to the x-axis
θ |
= |
π−π3 |
|
|
= |
3π3−π3 |
|
|
= |
2π3 |
θ |
= |
π+π3 |
|
|
= |
3π3+π3 |
|
|
= |
4π3 |
θ |
= |
2π−π3 |
|
|
= |
6π3−π3 |
|
|
= |
5π3 |
θ=π3,2π3,4π3,5π3
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Question 2 of 3
Given that 0≤θ≤2π, solve for θ
sin2θ=1
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1.
π2,3π2
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2.
π2,π,3π2,2π
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3.
π4,3π4
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4.
π4,3π4,5π4,7π4
Incorrect
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First, simplify the given equation
sin2θ |
= |
1 |
√sin2θ |
= |
√1 |
Get the square root of both sides |
sinθ |
= |
±1 |
In the unit circle. mark out the coordinates on the horizontal and vertical axis
Keep in mind that the radius of a unit circle is 1
Now, recall that in a coordinate, (x,y)=(cosθ,sinθ)
The y value in these coordinates match the value of sinθ. Therefore, θ=π2,3π2
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Question 3 of 3
Given that 0≤θ≤2π, solve for θ
cosθ=−1
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In the unit circle. mark out the coordinates on the horizontal and vertical axis
Keep in mind that the radius of a unit circle is 1
Now, recall that in a coordinate, (x,y)=(cosθ,sinθ)
The x value in π the value of cosθ. Therefore, θ=π