Solve the following equations giving any roots in terms of pi in the .......?
interval -2pi ≤ 0 ≤ 2pi.
a) 2cos²Θ + sin²Θ = 0
b) 2cos²Θ + sin²Θ = 1
c) 2cos²Θ + sin²Θ = 2
First convert cos^2 theta to 1 - sin^2 theta. Then treat sin^2 as a new variable x and solve for that using quadratic equation procedures.
I will do the last one for you
2cos²Θ + sin²Θ = 2
2cos²Θ + 1 - cos²Θ = 2
cos²Θ = 1
cosΘ = ±1
in degrees, Θ = 0º or 180º or 360º
in radians: Θ = 0 pi or 2pi
To solve these equations and find the roots in terms of π within the given interval, we can apply trigonometric identities and algebraic manipulation.
a) 2cos²Θ + sin²Θ = 0:
First, let's simplify the equation using the Pythagorean identity: sin²Θ + cos²Θ = 1.
Rearranging the equation, we get:
2cos²Θ = -sin²Θ
Dividing both sides by cos²Θ, we have:
2 = -tan²Θ
Taking the square root of both sides, we obtain:
√2 = ±√(-tan²Θ)
Since tan is undefined for certain values of Θ, the expression -tan²Θ is also undefined. Therefore, there are no roots in terms of π within the given interval for this equation.
b) 2cos²Θ + sin²Θ = 1:
Similarly, using the Pythagorean identity, we have:
2cos²Θ = 1 - sin²Θ
Substituting this value back into the equation, we get:
1 - sin²Θ + sin²Θ = 1
The equation simplifies to:
1 = 1
Since this equation is true for all values of Θ, there are infinite roots in terms of π within the given interval for this equation.
c) 2cos²Θ + sin²Θ = 2:
Again, using the Pythagorean identity, we have:
2cos²Θ = 2 - sin²Θ
Substituting this value back into the equation, we get:
2 - sin²Θ + sin²Θ = 2
The equation simplifies to:
2 = 2
Just like the previous equation, this equation is true for all values of Θ. Therefore, there are infinite roots in terms of π within the given interval for this equation as well.
In conclusion:
a) There are no roots in terms of π in the interval -2π ≤ Θ ≤ 2π.
b) There are infinite roots in terms of π in the interval -2π ≤ Θ ≤ 2π.
c) There are infinite roots in terms of π in the interval -2π ≤ Θ ≤ 2π.