Trigonometric Equation: 2sin(2x) = 2cos(x)

AI Thread Summary
The discussion centers on solving the trigonometric equation 2sin(2x) = 2cos(x). The initial attempts led to the equations 2cosx(sinx-1)=0 and 2cosx=0, with incorrect solutions of pi/2 and 3pi/2. Participants clarified that the correct approach involves finding solutions for sinx=1/2 and cosx=0, leading to arcsin(1/2) and arccos(0). After reviewing notes, the correct solutions were determined to be pi/6, 5pi/6, pi/2, and 3pi/2 within the domain of [0, 2pi]. The discussion emphasizes the importance of correctly identifying the values of x in trigonometric equations.
GuN
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Homework Statement



2 sin(2x) = 2 cos(x)

Homework Equations





The Attempt at a Solution



4sinxcosx-2cosx=0

2(2sinxcosx-cosx)=0

2cosx(sinx-1)=0

2cosx=0 sinx=1


And then I did the math and got pi/2 and 3pi/2 , but that's apparently wrong.
 
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I get cosx * (2*sinx - 1) = 0 for your second to last step. Does that help?
 
I still get pi/2 and 3 pi/2.

>_<

I'll try consulting my notes on how to properly calculate cosx=0 and sinx=1/2.
 
GuN said:
I still get pi/2 and 3 pi/2.
How? You know that ##\sin (\pi/2) = 1## and ##\sin (3\pi/2) = -1##. So obviously neither of those satisfy ##\sin x = 1/2##.
 
arcsin (1/2) = x1
arccos (0) = x2

You have done everything correctly, now you need to figure out what the values for x are. This is not a linear equation, there are 2 series of solutions.

The system I showed you will not be your final answer.
 
Right, I double checked the notes and found I used the wrong radians.

I redid it and got (for the domain of [0, 2pi] ) pi/6, 5pi/6, pi/2 and 3pi/2.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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