How Do You Solve the Equation \( \sqrt{3}\cos(x) + \sin(x) = 1 \)?

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To solve the equation \( \sqrt{3}\cos(x) + \sin(x) = 1 \), the initial approach involved transforming it into a quadratic form, but there were errors in the substitution of \(\sin^2(x)\) and maintaining the equation balance. Correcting these errors leads to the factorization \( \cos(x)(4\cos(x) + 2\sqrt{3}\sin(x)) = 0 \), yielding solutions where \( \cos(x) = 0 \) at \( x = 90^\circ \) or \( 270^\circ \). Further analysis suggests using the identity \( \tan(x) = \sin(x)/\cos(x) \) to explore additional solutions, while cautioning against extraneous solutions introduced by squaring both sides. The discussion emphasizes that there are infinitely many solutions due to periodicity, and checking solutions in the original equation is crucial.
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Homework Statement


Hi, I need to solve for x:
\sqrt{3}cos(x)+sin(x)=1


Homework Equations





The Attempt at a Solution


3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Sinx)^{2}-1=0
3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Cosx)^{2}=
4(Cosx)^{2}+2\sqrt{3}SinxCosx=0
Cosx(4Cosx+2\sqrt{3}Sinx)=0
\therefore Cosx=0 x=90 or 270.
OR
4Cosx=-2\sqrt{3}Sinx
I wasn't sure how to work out that last bit.
Hope you guys can help.
Frenzal
 
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frenzal_dude said:

Homework Statement


Hi, I need to solve for x:
\sqrt{3}cos(x)+sin(x)=1


Homework Equations





The Attempt at a Solution


3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Sinx)^{2}-1=0
3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Cosx)^{2}=
In the line above you replaced sin^2(x) - 1 with cos^2(x). That should be -cos^2(x).

Also, you lost the 0 on the right-hand side.
frenzal_dude said:
4(Cosx)^{2}+2\sqrt{3}SinxCosx=0
Cosx(4Cosx+2\sqrt{3}Sinx)=0
\therefore Cosx=0 x=90 or 270.
OR
4Cosx=-2\sqrt{3}Sinx
I wasn't sure how to work out that last bit.
Hope you guys can help.
Frenzal
Use the identity that sin(x)/cos(x) = tan(x). You will first need to fix the error noted above, though.

Also, be sure to check your solutions in the original equation. By squaring both sides, you might be introducing extraneous solutions: solutions of your squared equation that are not solutions of the original equation.

One other thing. Since there are no restrictions on x, there are going to be an infinite number of solutions. For example, if x = pi/6 were to turn out to be a solution, then pi/6 + n*2pi, n = 0, +/-1, +/-2, ... would represent all such solutions.
 
divide both sides of the eqn by 2.
& then u can write that as
sin60*cosx + cos60*sinx = 1/2
sin(x + 60) = sin(30) = sin(150)
x = -30, 90

& then substitute & check which is correct
 
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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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