# Trigonometrysolving trig equation cos(x)=sin(x) + 1/√3

#### sp3

##### New member
Hello, I'm trying to solve the following equation : cos(x)=sin(x) + 1/$$\displaystyle \sqrt{3}$$
in order to find cos(x)3 - sin3(x) = ?

i tried to slove for sin(x) using sin2 + cos2 =1 replacing cos(x) by the first equation and i end up with a second degree polynomial using x = sin(x) and there are 2 solutions, it seems off....
btw i used the (a+b)2 identity for (sin(x) + 1/$$\displaystyle \sqrt{3}$$)2

if anyone could help me i thank you in advance!

#### HallsofIvy

##### Well-known member
MHB Math Helper
Okay, I presume you wrote $$\sqrt{1- sin^2(x)}= sin(x)+ \frac{1}{\sqrt{3}}$$ and then, squaring both sides, $$1- sin^2(x)= sin^2(x)+ \frac{2}{\sqrt{3}}sin(x)+ \frac{1}{3}$$.

Then we have $$sin^2(x)+ \frac{1}{\sqrt{3}}sin(x)- \frac{1}{3}= 0$$.

By the quadratic formula, $$sin(x)= \frac{-\frac{1}{\sqrt{3}}\pm\sqrt{\frac{1}{3}+ \frac{4}{3}}}{2}$$$$= \frac{-1\pm \sqrt{5}}{2\sqrt{3}}$$.

Yes, there are two solutions (between 0 and $$2\pi$$). Numerically, sin(x)= 0.3568 so x= 0.3649 and sin(x)= -0.9342 so x= -1.2059 to four decimal places.

#### sp3

##### New member
thank you! i find the same thing but if i have to solve cos(x)3- sin(x)3 how do i get to one answer with this $$\displaystyle \pm$$ ? the answer is $$\displaystyle \frac{4}{3\sqrt{3}}$$. I thought about using the identity
a3-b3 = (a-b)(a2+2ab+b2) as an alternative method with a=sin(x)+$$\displaystyle \frac{1}{\sqrt{3}}$$ and b=sin(x) . I find this :

$$\displaystyle \frac{12sin(x)^2+\sqrt{3}*4sin(x)+1}{3\sqrt{3}}$$

Idk how to further develop. In the identity I factorized $$\displaystyle (sin+\frac{1}{\sqrt{3}})^2$$, found in (a-b) sinx-sinx =0 so there’s $$\displaystyle \frac{1}{\sqrt{3}}$$ left. any help is welcome!

Btw people here are so quick to answer this website is bomb, thank you really! Appreciate it

Last edited:

##### Well-known member
Hello SP3.
you do not need $\sin\,x$ or $\cos\,x$
you are given
$\cos\,x - \sin\,x = \frac{1}{\sqrt{3}}$
$\cos^3 x - \sin ^3 x = (\cos\,x - \sin\,x)^3 + 3 (\cos\,x - \sin\,x)(\cos\,x \sin\,x)\cdots(1)$
so you need to evaluate $\cos\,x \sin\,x$
$\cos\,x - \sin\,x = \frac{1}{\sqrt{3}}$
square both sides to get $1 - 2\cos\,x \sin\,x = \frac{1}{3}$
or $2\cos\,x \sin\,x = \frac{2}{3}$
or $\cos\,x \sin\,x = \frac{1}{3}$
you can put the value of $\cos\,x \sin\,x = \frac{1}{3}$ and $\cos\,x - \sin\,x = \frac{1}{\sqrt{3}}$ in (1) to get
$\cos^3 x - \sin ^3 x = (\cos\,x - \sin\,x)^3 + 3 (\cos\,x - \sin\,x)(\cos\,x \sin\,x)$
$= \frac{1}{3\sqrt{3}} + 3 \frac{1}{\sqrt{3}} \frac{1}{3}$
$=\frac{1+3}{3\sqrt{3}}$
$=\frac{4}{3\sqrt{3}}$