Solving a quadratic trigonometric equation?

Question

The equation is $6 \cos^2x+\cos x=1$,

My work:

$6x^2+x-1=0$

$(3x-1)(2x+1)$

$3x-1=0 ∨ 2x+1=0$

$x=\frac{1}{3} ∨ x= \frac{-1}{2}$

But I do not know how to progress further.

Answer 1

It should be: $$\eqalign{6\cos^2x+\cos x-1=0 &\iff (3\cos x-1)(2\cos x+1)=0 \\ &\iff 3\cos x-1=0,\;\; 2\cos x+1=0 \\ &\iff\cos x=\tfrac13,\;\; \cos x=-\tfrac12.}$$ Can you take it from here?

Answer 2

Just solve $\cos(x)=1/3$ and $\cos(x)=-1/2$ and you're done.

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Tip: when making a substitution, use a different variable to the one you're already using.

e.g. in your case, you let $$\color{red}{x=\cos(x)},$$ whereas you should let $$\color{green}{\theta=\cos(x)} .$$

Answer 3

I'd set a different variable to $\cos(x)$ other than $x$ so you don't get them mixed up (like $\alpha$ or something).

You can see from your working that

$$\cos(x)=\frac{1}{3} \Rightarrow x= \cos^{-1}\left(\frac{1}{3}\right)$$ $$\therefore \quad x \approx 1.23+2\pi n, \quad x \approx5.05+2\pi n$$

$$\cos(x)=-\frac{1}{2} \Rightarrow x= \cos^{-1}\left(-\frac{1}{2}\right)$$ $$\therefore \quad x =\frac{2\pi}{3}+2\pi n, \quad x=\frac{4\pi}{3}+2\pi n$$

Answer 4

I would not use the same letter, $x$, to refer to two different things. $$ 6\cos^2 x + \cos x - 1 = 0 $$ $$ 6u^2 + u - 1 = 0 $$ $$ (3u-1)(2u+1)=0 $$ $$ u = \frac 1 3\text{ or } u = \frac{-1}{2} $$ $$ \cos x=\frac 1 3 \text{ or }\cos x = \frac{-1}2 $$ $$ \Big(x = \left(\arccos\frac 1 3\right) + 2\pi n \text{ or } \left(\pi-\arccos\frac 1 3\right) + 2\pi n \Big)\text{ or }\Big( x = \frac{2\pi}3+2\pi n \text{ or }x=\frac{4\pi}3+2\pi n \Big) $$

Answer 5

It is preferable to change the variable so let $t=\cos x$ and we solve the quadratic equation $$6t^2+t=1$$ and we find $t=\frac13$ and $t=-\frac12$. Now $$\cos x=\frac 13\iff \left(x=\arccos\left(\frac13\right)+2k\pi\right)\lor\left( x=2k\pi-\arccos\left(\frac13\right)\right)$$ and $$\cos x=-\frac12\iff \left(x=\frac{2\pi}3+2k\pi\right)\lor\left( x=2k\pi-\frac{2\pi}3\right)$$