Algebraic solution of Trig equation.

[synthetic.]

The question as is written :

Solve for x

5cosx-2= -0.72 ( 0 $$\leq$$ x $$\leq$$ 2$$\Pi$$ )

(The answer given is "1.31, 4.97".)


So, i done what seemed natural.


5cosx-2=-0.72

5cosx = 1.28
cosx = 1.28/5
cosx = 0.256


Firstly, I'm not even sure what unit my answer is supposed to be in. Since x is defined as Radians in the question, my answer of .256 = .256pi Radians (.256 of 180)?
Which should give solutions in the 1st & 4th quadrants.

All of the questions until (there are a bunch similar to this) these ones have been in the form of, for example, 4=5 - tan2x (x in radians), which, as there are no decimals in the solution work out easy (pi/8, 5pi/8).. . why, if x is defined in Radians in the question, are the answers in decimals? I don't understand the principle of this when decimals are involved for some reasons.

 

[rocomath]

$$x=\cos^{-1}{\left(\frac{2-0.72}{5}\right)}$$

Should be in radians.

Find the values in the 1st and 4th quadrant.

 

[rock.freak667]

4=5 - tan2x (x in radians), which, as there are no decimals in the solution work out easy (pi/8, 5pi/8).. . why, if x is defined in Radians in the question, are the answers in decimals? I don't understand the principle of this when decimals are involved for some reasons.

Well if 5-tan2x=4 then that is the same as tan2x=1,right? and arctan(1)=pi/4

The only reason why are usually get your answers in terms of pi are most likely due to the fact that the questions you do end up with a trig function being equal to a standard number.

For example, $sin3x=\frac{\sqrt{3}}{2} \Rightarrow 3x=sin^{-1}(\frac{\sqrt{3}}{2})$

and $sin^{-1}(\frac{\sqrt{3}}{2})=\frac{\pi}{3}$