Solve.
2sin(2ƒÆ) + �ã3 = 0
interval [0, 2ƒÎ)
How do I start this? I feel like maybe Trig Identities, but I'm not really sure.
I will assume your equation says something like
2sin(2πØ) + √3 = 0
if so, then
2sin(2πØ) = -√3
sin(2πØ) = -√3/2
The 'reference angle' is π/3 (I know sin π/3 = sin60° = √3/2)
the sine is negative in quads III and IV
so 2πØ = π+π/3 or 2πØ = 2π - π/3
Ø = 2/3 or 5/6
45
To solve the equation 2sin(2ƒÆ) + �ã3 = 0 in the interval [0, 2ƒÎ), you are correct that trigonometric identities will be useful. Specifically, you can use the double-angle identity for sine, which states that sin(2ƒÆ) = 2sin(ƒÆ)cos(ƒÆ).
Here's how you can solve it step-by-step:
Step 1: Rewrite the equation using the double-angle identity:
2(2sin(ƒÆ)cos(ƒÆ)) + �ã3 = 0
Step 2: Simplify the equation:
4sin(ƒÆ)cos(ƒÆ) + �ã3 = 0
Step 3: Factor out sin(Į):
sin(ƒÆ)(4cos(ƒÆ)) + �ã3 = 0
Step 4: Rearrange the equation:
4cos(ƒÆ)sin(ƒÆ) + �ã3 = 0
Step 5: Since the equation involves sine and cosine, it will be useful to convert it into a single trigonometric function. Using the identity sin(2Į) = 2sin(Į)cos(Į), we can rewrite the equation as:
2sin(2ƒÆ) + �ã3 = 0
Step 6: At this point, you can solve the equation using algebraic methods. Subtract �ã3 from both sides:
2sin(2ƒÆ) = -�ã3
Step 7: Divide both sides by 2:
sin(2ƒÆ) = -�ã3/2
Step 8: Now, you need to find the solutions for sin(2ƒÆ) = -�ã3/2 within the given interval [0, 2ƒÎ). To solve this, it will be helpful to know the values of sin for different angles.
Step 9: In the given interval [0, 2ƒÎ), identify the angles where sin(2ƒÆ) = -�ã3/2. These angles can be found by using the inverse sine function or by knowing the trigonometric ratios for specific angles. In this case, 2ƒÆ = 7ƒÑ/6 or 11ƒÑ/6.
Step 10: Convert the angles back to Į by dividing by 2:
Į = 7Ą/12 or 11Ą/12.
Therefore, the solutions to the equation 2sin(2ƒÆ) + �ã3 = 0 in the interval [0, 2ƒÎ) are ƒÆ = 7ƒÑ/12 and ƒÆ = 11ƒÑ/12.