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I tried to solve this equation, but my solution is wrong and I don't understand why. the answer in the book is: $x = \pm60+180k$. my answer is: $x= \pm60+360k$. please help :)

3cos(x)^2 = sin(x)^2
3cos(x)^2 = 1 - cos(x)^2

t = cos(x)^2

3t=1-t
4t=1
t=1/4

cos(x)^2 = 1/4
cos(x) = 1/2
cos(x) = cos(60)

x = +-60+ 360k

edited:

cos(x) = -1/5
cos(x) = cos(120)
x = +-120 + 360k

I still don't get the answer in the book

Michael Hardy
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Silas2033
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3 Answers3

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\begin{align*} 3\cos^2 x & = \sin^2x\\ 3 & = \frac{\sin^2x}{\cos^2x}\\ \tan^2x & = 3\\ \tan x & = \sqrt{3}\\ \tan x & = \tan 60 \end{align*}

From there, you can use the tangent equation:

$$x= \pm 60+180k$$

N. F. Taussig
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akusaja
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    You can see how I typeset your work by right-clicking on an equation, then selecting Show Math Commands As LaTeX Commands. Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Sep 29 '15 at 21:12
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Since $\sin^2 x=1-\cos^2 x$ the given equation is equivalent to \begin{align} 3\cos^2 x&=1-\cos^2 x\\ \iff \quad \cos^2 x&=\frac{1}{4}\\ \iff \cos x&\in\left\{-\frac{1}{2},\frac{1}{2}\right\} \end{align} Hence $$x=180^{\circ}\cdot k \pm 60^{\circ}$$ where $k$ is an integer number.

Silas2033: Notice that in this problem $\cos x$ can be either $\frac{1}{2}$ or $-\frac{1}{2}$, then you must regard two cases:

1) $\cos x = \cos y$ giving us $x=\pm y +360^{\circ}k=\pm y +180^{\circ}(2n)$ and

2) $\cos x = -\cos y$ giving us $x=180^{\circ}-(\pm y +360^{\circ}k)=\mp y - 180^{\circ}(2k-1)$.

Ángel Mario Gallegos
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Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$

$$3\cdot\dfrac{1+\cos2A}2=\dfrac{1-\cos2x}2\iff\cos2x=-\dfrac12=-\cos60^\circ$$

and as $\cos(180^\circ-y)=-\cos y$

$$\cos2x=\cos(180^\circ-60^\circ)\iff2x=360^\circ n\pm120^\circ$$ where $n$ is any integer

lab bhattacharjee
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