Hey Tina🤭🌹, good to see you again.
Here are the solutions to the problems.
11. Solve for θ in the interval 0≤θ≤2π:
2sinθcosθ=1
Step 1: Use the double angle identity sin(2θ)=2sinθcosθ.
sin(2θ)=1
Step 2: Find the values for 2θ where sin(2θ)=1.
The general solution for sinx=1 is x=2π+2nπ, where n is an integer.
So, 2θ=2π+2nπ.
Step 3: Solve for θ.
θ=4π+nπ
Step 4: Find the values of θ in the interval 0≤θ≤2π.
For n=0: θ=4π
For n=1: θ=4π+π=45π
For n=2: θ=4π+2π=49π, which is outside the interval.
The solutions are 4π and 45π.
θ=4π,45π
12. Solve for θ in the interval 0∘≤θ≤360∘:
sin2θ=cosθ
Step 1: Use the double angle identity sin(2θ)=2sinθcosθ.
2sinθcosθ=cosθ
Step 2: Rearrange the equation to set it to zero.
2sinθcosθ−cosθ=0
Step 3: Factor out cosθ.
cosθ(2sinθ−1)=0
Step 4: Set each factor equal to zero and solve for θ.
Case 1: cosθ=0
For 0∘≤θ≤360∘, cosθ=0 when θ=90∘ or θ=270∘.
Case 2: 2sinθ−1=0
2sinθ=1
sinθ=21
For 0∘≤θ≤360∘, sinθ=21 when θ=30∘ (in Quadrant I) or θ=180∘−30∘=150∘ (in Quadrant II).
The solutions are 30∘,90∘,150∘,270∘.
θ=30∘,90∘,150∘,270∘
13. Solve for θ in the interval 0≤θ≤2π:
cos2θ+sinθ=0
Step 1: Use the double angle identity cos2θ=1−2sin2θ.
1−2sin2θ+sinθ=0
Step 2: Rearrange the equation into a standard quadratic form.
−2sin2θ+sinθ+1=0
Multiply by −1 to make the leading coefficient positive.
2sin2θ−sinθ−1=0
Step 3: Let x=sinθ and solve the quadratic equation 2x2−x−1=0.
Factor the quadratic equation:
(2x+1)(x−1)=0
Step 4: Substitute back sinθ and solve for θ.
Case 1: 2sinθ+1=0
2sinθ=−1
sinθ=−21
For 0≤θ≤2π, sinθ=−21 in Quadrant III and Quadrant IV.
The reference angle is 6π.
In Quadrant III: θ=π+6π=67π
In Quadrant IV: θ=2π−6π=611π
Case 2: sinθ−1=0
sinθ=1
For 0≤θ≤2π, sinθ=1 when θ=2π.
The solutions are 2π,67π,611π.
θ=2π,67π,611π
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