Answer :
Answer:
Absolutely, let's find all solutions for both equations:
1. √2 sin theta + 1 = 0
We can solve this equation by isolating sin(theta) and then using the inverse sine function (sin^-1 or arcsin). However, it's important to remember that the inverse sine function only outputs values within the range of -π/2 to π/2 (or -90° to 90°) even though sine might have multiple solutions for a specific value within its range. To find all possible solutions, we need to consider adding or subtracting multiples of 2π (or 360°) because the sine function repeats its values every 2π (or 360°).
Steps to solve:
Isolate sin(theta): sin(theta) = -(1 / √2)
Inverse sine function: theta = sin^-1( -(1 / √2) )
Finding all solutions:
The inverse sine function will give us one principal solution in the range of -π/2 to π/2:
theta ≈ -π/4 (or ≈ -45°)
Considering periodicity:
Since sine repeats every 2π, other solutions include:
-π/4 + 2π ≈ 7π/4 (or -45° + 360° ≈ 315°)
-π/4 - 2π ≈ -5π/4 (or -45° - 360° ≈ -405°)
2. 2cos^2 theta - 1 = 0
This equation can be rewritten to solve for cos(theta).
Steps to solve:
Add 1 to both sides: 2cos^2 theta = 1
Divide both sides by 2: cos^2 theta = 1/2
Take the square root of both sides (be aware of positive and negative): cos(theta) = ±√(1/2) = ±(1/√2)
Finding all solutions:
Now we can find the solutions using the inverse cosine function for both positive and negative values:
cos^-1(1/√2) ≈ π/4 (or 45°)
cos^-1(-1/√2) ≈ 3π/4 (or 225°)
Considering periodicity:
As with sine, cosine also repeats every 2π. Therefore, all solutions include adding or subtracting multiples of 2π:
π/4 + 2πn ≈ (45° + 360°n), where n is any integer
3π/4 + 2πn ≈ (225° + 360°n), where n is any integer
Summary:
Equation 1: theta ≈ -π/4 + 2πn (all multiples of π/4 shifted by -π/4, which is approximately -45° + 360°n)
Equation 2: theta ≈ π/4 + 2πn and theta ≈ 3π/4 + 2πn (all multiples of π/4 shifted by π/4 and 3π/4, which are approximately 45° + 360°n and 225° + 360°n)
