solve the equation for θ ( 0 ≤ θ ≤ 2π )?

Question

2sin^2θ = 1

tan^2θ - tanθ = 0


plz help i dont understand how 2 do these!! plz show steps thanks!!!!

Answer 1

2sin^2(θ) = 1

Divide both sides by 2:
sin^2(θ) = 1/2

Take sqrt of both sides:
sin(θ) = ±sqrt(1/2)
sin(θ) = ±1 / sqrt(2)
sin(θ) = ±sqrt(2) / 2

Using the unit circle:
θ = π/4, 3π/4, 5π/4, 7π/4

You can verify these answers by plugging them back into the original equation (they all work).


tan^2(θ) - tan(θ) = 0

Factor out a tan(θ):
tan(θ) [tan(θ) - 1] = 0

tan(θ) = 0
θ = 0, π, 2π

tan(θ) - 1 = 0
tan(θ) = 1
θ = π/4, 5π/4

So θ = 0, π/4, π, 5π/4, 2π

Once again, you can verify the answers.

Answer 2

1) First divide by 2
sin^2θ = 1/2
Now take the square root
sin θ = 1/sqrt(2) = sqrt(2) / 2
The θ between 0 and 2π that satisfies this are:
π/4 and 3π/4 (look at a graph of y = sin x and see where they cross the line y = sqrt(2) / 2

2) tan^2θ - tanθ = 0
Factor out a tanθ:
tanθ (tanθ - 1) = 0
Thus the solutions are when
tanθ = 0 and tanθ = 1
a) tanθ = 0 when sinθ = 0, which is when
θ =0, π, 2π
b) tanθ = 1 when sinθ = cosθ, which is when
θ =π/4, 5π/4 (has to be in the first and third quadrants for the sin and cos to be both positive or both negative)
Thus your final answer is θ =0, π/4, π, 5π/4, 2π

Answer 3

I'm going to use x instead of theta, ok?

1st one:

Divide by 2:

sin^2x = 1/2

Take the square root of both sides:
sin(x) = +/- sqrt2/2

Using unit circle, there are 4 places that solve for x:

x = pi/4, 3pi/4, 5pi/4, and 7pi/4

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2nd prob:

Factor out tan(x): tan(x){(tan(x) -1)} = 0

So tan (x) = 0 or tan(x) = 1

tan(x) = 0 when sin(x) = 0, so that is at x = 0, pi, and 2pi

tan(x) = 1 when sin(x) = cos(x) , so x = pi/4 or 5pi/4

Answer 4

sin^2 t = 1/2 so sin t =+-1/2sqrt(2)
t=pi/4 ,3pi/4,5pi/4 and 7pi/4
tan t(tan t-1) tan t= 0 t= 0 ,pi and 2pi
tant = 1 t= pi/4 and 5pi/4

Answer 5

i just use x for theta

2sin^2(x) = 1

divide 2 for both sides
sin^2(x) = 1/2

take a square root
sin(x) = +sqrt(1/2) or -sqrt(1/2)

sin(x) = sqrt(1/2)
x = pi/4 and 3pi/4

sin(x) = -sqrt(1/2)
x = 7pi/4 and 5pi/4


tan^2(x) - tan(x) = 0

take out tan(x)
tan(x) (tan(x) - 1) = 0

tan(x) = 0
x = 0 and 2pi

tan(x) - 1 = 0
tan(x) = 1
x = pi, pi/4 and 5pi/4

Answer 6

θ=π/12
θ=5π/12