Question 325704: the book asks to find all solutions in the interval(0.2Pi).
the problem is:
1-sin t=sqroot3 cos t
What I did was bring the right side over to the left by subtracting it from both sides. (1-sin t)-(sqroot3 cos t)=0 using zero factor theory, then I would have: sin t= 1,the solution is Pi/2, and I'm not quite sure how to solve the other term , I thought I could divide the sqrt3 from both sides leaving cos t=0 which the answer for that would also be Pi/2. The answers in the back of the book are: 11Pi/6, and Pi/2. so I know I have one right but I can't figure out how to get the second answer.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Let's see, you started with:
\ =\ \sqrt{3}\,\cos(t))
And then you wrote:
\ -\ \sqrt{3}\,\cos(t)\ =\ 0)
Which is perfectly ok, in and of itself, but as we will soon see, not very helpful.
Then you really went astray. You tried to apply the Zero Factor Property.
But ) and ) are NOT factors that when multiplied together equal zero. The Zero Factor Property applies to a situation like  from which we can tell that either  or  . But your situation is  . All you can tell from that is that  -- but you already knew that from the original equation.
Start with the original equation and square both sides, applying FOIL to the LHS.
\ +\ \sin^2(t)\ =\ 3\,\cos^2(t))
Ewww! That's ugly! Have faith, Grasshopper...
Recall the Pythagorean Identity,  , which can also be written:  .
Then make the substitution into your equation:
\ +\ \sin^2(t)\ =\ 3\left(1\ -\ \sin^2{t}\right))
Distribute the 3 in the RHS, and then collect like terms with everything in the LHS:
\ -\ \sin(t)\ -\ 1\ =\ 0)
Let ) , then:
Ah Ha! Now it is starting to look like something familiar, no?
Factor:
(2u\ +\ 1)\ =\ 0)
NOW you can use the Zero Factor Property.
or
Substitute back:
\ =\ 1\ \Rightarrow\ t\ =\ \sin^{-1}\left(1\right))
or
\ =\ -\frac{1}{2}\ \Rightarrow\ t\ =\ \sin^{-1}\left(-\frac{1}{2}\right))
Look at the Unit Circle:
And recall that on the unit circle,  is represented by the  -coordinate.
In the interval [ , ], there is one angle that has a -coordinate of 1, namely  . There are 2 angles that have a -coordinate of  , namely  and  .
However, you need to check your answers, which is left as an exercise for the student. Submit each of the angle values to the original equation. 2 of them check, but one of them does not.
John
|
|
|
