Question

solve the equation 6sinθcosθ+4cos2θ=1, for 0<θ 11 years ago

Answer 1

you can use:
\[2\sin(\theta)\cos(\theta)=\sin(2 \theta) \\ \text{ so multiply on both sides 3 we have } \\ 6 \sin(\theta) \cos(\theta)=3 \sin(2 \theta) \\ \text{ so we have } 3 \sin(2 \theta)+4 \cos(2 \theta)=1 \\ \text{ Let } u=2 \theta \\ \text{ so since } 0< \theta< \pi \\ \text{ then } 0(2)<\theta(2)<\pi(2) \\ \text{ or} 0<2 \theta< 2 \pi \\ \text{ which means } \\ 0 <u < 2\pi \\ \\ \text{ so we want \to solve } \\ 3 \sin(u)+4 \cos(u)=1 \text{ for } 0<u<2\pi \\ \frac{3}{\sqrt{3^2+4^2}}\sin(u)+\frac{4}{\sqrt{3^2+4^2}} \cos(u)=\frac{1}{\sqrt{3^2+4^2}} \]
construction of triangle:

Answer 2

\[\sin(v)\sin(u)+\cos(v)\cos(u)=\frac{1}{\sqrt{9+16}} \\ \cos(v-u)=\frac{1}{\sqrt{25}} \\ \cos(v-u)=\frac{1}{5}\]

Answer 3

so you might want to find the angle v first given the triangle I have above

Answer 4

ohhh okay thank you, i didn't think of rearranging the trigonometric equation at first.

Answer 5

@freckles can this be written in a Rcos(v-u) form
i.e. 5cos(2θ- ? )
as this is what we have done previously in this topic