Solve each equation for 0 ≤ θ ≤ 2π, giving your answers in terms of π:
a. sin 2θ = sin θ
b. cos 2θ + sin θ = 0
c. sin2θ = √3cos θ
d. cos θ = sin θ sin 2θ
e. cos 2θ = cos θ



Answer :

Answer:

Step-by-step explanation:

A. :

Recall that sin 2Θ = 2sinΘcosΘ

So,

2sinΘcosΘ = sinΘ

cosΘ = [tex]\frac{sin(O)}{2sin(O)}[/tex] = 1/2

Recalling special right triangles or simply memorizing trigonometric values, cos (π/3)  = 1/2 or Θ = π/3 or 5π/3

B. :

Recall that cos2Θ = 1 - 2sin^2(Θ) = 2cos^2(Θ) - 1

For our cases let's rewrite cos2Θ in the same terms as sinΘ.

1 - 2sin^2(Θ) + sinΘ = 0

-1 + 2sin^2(Θ) - sinΘ = 0  

(multiplied both sides by -1 to make sin^2 positive)

2sin^2(Θ) - sin(Θ) - 1 = 0  (rearranged)

Hey, this looks like a quadratic equation!

We can make the factoring easier by assigning sinΘ = x,

2x^2 - x - 1 = 0   (now that's better!)

(2x+1)(x-1) = 0

x = -1/2 or 1

sinΘ = -1/2 or sinΘ = 1

So, Θ = 11π/6 or π/2

C. :

Plugging the formula of sin2Θ into the equation given, we get,

2sinΘcosΘ = √3cosΘ

2sinΘ = [tex]\frac{\sqrt{3}cosO)}{cos(O)}[/tex]

2sinΘ = √3

sinΘ = (√3)/2

So, Θ = π/3 or 2π/3

D. :

Again, plugging the formula of sin2Θ into the equation given, we get,

cosΘ = sinΘ(2sinΘcosΘ)

cosΘ = 2sin^2(Θ)cosΘ (sinΘ*sinΘ is like x*x = x^2)

cosΘ/cosΘ = 2sin^2(Θ)   (divide both sides by cosΘ)

1 = 2sin^2(Θ)

1/2 = sin^2(Θ)  (divide both sides by 2)

√(1/2) = sinΘ  (square root both sides to get rid of the ^2 on the right side)

√2/2 = sinΘ

So, Θ = π/4 or 3π/4

E.  :

Plugging the formula of cos2Θ into the given equation, we get,

2cos^2(Θ) - 1 = cosΘ

2cos^2(Θ) - cosΘ - 1 = 0    (subtract cosΘ both sides)

cosΘ = x

2x^2 - x - 1 = 0   (same quadratic as B)

cosΘ = -1/2 or cosΘ = 1

So, Θ = 2π/3 or Θ = 0/2π  

(we're dealing with the cosine function so, the values Θ will be different from B's)

Let me know if you have any questions!