Answer :
Answer:
Step-by-step explanation:
Given that a trignometric equation is as follows
[tex]2cos(3t) =1\\[/tex]
Divide by 2
[tex]cos(3t) =0.5\\[/tex]
b) 3t would be either in I quadrant or IV quadrant in the interval [0, 2π).
[tex]3t=\frac{\pi}{3} ,2\pi- \frac{\pi}{3} \\t =\frac{\pi}{9} , \frac{5\pi}9}[/tex]
a) To find general solution
General solution is
±[tex]2k\pi[/tex]±[tex]\frac{\pi}{3}[/tex]
Answer:
a) [tex]\theta=\pm \frac{\pi}{9},\frac{2}{3}\pi\pm \frac{\pi}{9},\frac{4}{3}\pi\pm \frac{\pi}{9},....[/tex]
b) [tex]\theta=\frac{\pi}{9}[/tex]
Step-by-step explanation:
Given : Equation [tex]2\cos(3\theta)=1[/tex]
To find : (a) Find all solutions of the equation.
(b) Find the solutions in the interval [0, 2π).
Solution :
Solving equation,
[tex]2\cos(3\theta)=1[/tex]
[tex]\cos(3\theta)=\frac{1}{2}[/tex]
We know, [tex]\cos(\frac{\pi}{3})=\frac{1}{2}[/tex]
[tex]\cos(3\theta)=\cos(\frac{\pi}{3})[/tex]
The general solution is
[tex]3\theta=2k\pi \pm \frac{\pi}{3}[/tex]
[tex]\theta=\frac{2}{3}k\pi \pm \frac{\pi}{9}[/tex]
(a) All solutions of the equation
i.e. for k=0,1,2,3....
For k=0, [tex]\theta=\pm \frac{\pi}{9}[/tex]
For k=1, [tex]\theta=\frac{2}{3}\pi\pm \frac{\pi}{9}[/tex]
For k=2, [tex]\theta=\frac{4}{3}\pi\pm \frac{\pi}{9}[/tex]
So, the solutions are [tex]\theta=\pm \frac{\pi}{9},\frac{2}{3}\pi\pm \frac{\pi}{9},\frac{4}{3}\pi\pm \frac{\pi}{9},....[/tex]
(b) The solutions in the interval [0, 2π).
As cos is positive in only first quadrant.
So, solution exist from [0, 2π) is when
[tex]\cos(3\theta)=\cos(\frac{\pi}{3})[/tex]
[tex]3\theta=\frac{\pi}{3}[/tex]
[tex]\theta=\frac{\pi}{9}[/tex]