Solve Trigonomic Equation: 5cos2x + cosx + 2 = 0

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The discussion focuses on solving the trigonometric equation 5cos2x + cosx + 2 = 0 within the interval 0 ≤ x ≤ 360. The user attempts to substitute for cos2x using various identities but becomes stuck after simplifying to a quadratic equation in terms of cosx. Another participant suggests substituting cosx with y to solve the quadratic equation, emphasizing that this method is standard for such problems. The original poster realizes their oversight in recognizing the quadratic form and expresses gratitude for the clarification. They conclude by noting they will now calculate the specific angles needed for the solution.
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Homework Statement



Solve the following trigonomic equation for 0<=x360<= :
5cos2x + cosx + 2 = 0

Homework Equations



Not sure what goes here but I've been trying to substitute these -
cos2x = cos^2x - sin^2x
cos2x = 2cosx^2 - 1 - I think this is the correct one to use.
cos2x = 1 - sin^2x

The Attempt at a Solution



5cos2x + cosx + 2 = 0
5(2cos^2x-1) + cosx + 2 = 0
10cos^2x + cosx - 3 =0
cosx (10cosx +1) - 3 = 0
? I get stuck here

Am I going about this the right way, or am I completely lost? I think I'm doing it right, I'm just not sure how to continue.

Any help appreciated.
 
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From here: 10cos^2x + cosx - 3 =0, make the substitution cosx=y, and solve the quadratic equation for y. Then, say you have solutions y=a,b, you can then solve the equations a=cosx, b=cosx to obtain all values of x.
 
Hi. Thanks for the reply.

I've never been taught how to do that, so is there another way of doing it?

Thanks again,
Mark
 
markyp23 said:
Hi. Thanks for the reply.

I've never been taught how to do that, so is there another way of doing it?

Thanks again,
Mark

You have a quadratic in cosx. Now, the usual ways to solve the equation are to either use the quadratic equation, or to spot some factors and factorise the expression. I used the substitution y=cosx to simplify the algebra throughout the calculation of the roots of the quadratic equation.

Perhaps your teacher hasn't explicitly taught you this, since there isn't really anything to be taught here! He expects you to be able to solve a quadratic equation, and here you have one in cosx.
 
Thank you!

I just figured out what you were saying. I didn't see it as a quadratic equation at all - oh dear (exam in 4 day, haha).

Figured it out, now just need to calculate the actual angles invovled. Thanks again.

Mark
 
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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