trig

[carebear]

1. Solve tanx = 2 sinx [0,2pi]

Is this correct?

multiply equation by cosx to get sinx - 2sinxcosx =0
sinx(1-2cosx)=0
sinx=0, cosx=0.5

x= 0, pi, 2pi x= pi/3, 5pi/3

2. Don't know where to start....

Solve cosx + 1 = sqrt3 sin x

...Please help

 

[galactus]

2. Don't know where to start....

Solve cosx + 1 = sqrt3 sin x

...Please help

\(\displaystyle cos(x)+1-\sqrt{3}sin(x)=0\)

Rewrite as:

\(\displaystyle 2cos(x+\frac{\pi}{3})+1=0\)

Now, continue.

 

[carebear]

I don't understand how you rewrote that....could you please explain that broken down....

 

[Deleted member 4993]

I don't understand how you rewrote that....could you please explain that broken down....

\(\displaystyle A * cos(x) - B* sin(x) = \ \sqrt{A^2+B^2}\cdot \left (\frac{A}{\sqrt{A^2+B^2}}\cdot cos(x) \ - \ \frac{B}{\sqrt{A^2+B^2}}\cdot sin(x)\right )\)

let

\(\displaystyle \frac{A}{\sqrt{A^2+B^2}} \ = \ cos(\theta)\)

then

\(\displaystyle \frac{B}{\sqrt{A^2+B^2}} \ = \ sin(\theta)\)

we get

\(\displaystyle A * cos(x) - B* sin(x) = \ \sqrt{A^2+B^2}\cdot [cos(\theta)\cdot cos(x) \ - \ sin(\theta)\cdot sin(x)] \ = \ \ \sqrt{A^2+B^2}\cdot [cos(\theta + x)]\)

In your case:

A = 1

B = ?3