Let theta be an acute angle such that sin2theta=sin3theta What is the measure of theta in degrees?

itsash Dec 9, 2023

#1**0 **

We're given that sin2θ=sin3θ where θ is an acute angle. Our goal is to find the measure of θ in degrees. Using double-angle identities

From the double-angle identity for sine, we have:

sin2θ=2sinθcosθ

Substituting this into the given equation, we get:

2sinθcosθ=sin3θ

Using another double-angle identity for sine:

$$ \sin 3\theta = 3\sin\theta - 4\sin^3\theta$$

Substituting this again, we get:

2sinθcosθ=3sinθ−4sin3θ

Simplifying:

2sinθcosθ−3sinθ+4sin3θ=0

Factoring sinθ out:

sinθ(2cosθ−3+4sin2θ)=0

Since θ is acute, sinθ\ge0. Therefore:

2cosθ−3+4sin2θ=0

This equation is equivalent to:

4sin2θ−2cosθ−3=0

We can use the quadratic formula to solve for cosθ:

cosθ=2⋅42±(−2)2−4⋅4⋅−3

cosθ=82±52

Discarding the negative solution because cosθ is positive for acute angles, we get:

cosθ = (2 + sqrt(52))/8

Using the inverse cosine function, we get:

θ = arccos((2 + sqrt(52)/8)

In degrees, the measure of θ is approximately:

θ=47.1∘

BuiIderBoi Dec 10, 2023