MHB Solving trig equation cos(x)=sin(x) + 1/√3

[sp3]

Hello, I'm trying to solve the following equation : cos(x)=sin(x) + 1/$$\sqrt{3}$$
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/$$\sqrt{3}$$)²

if anyone could help me i thank you in advance!

 

[HallsofIvy]

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]

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 $$\pm$$ ? the answer is $$\frac{4}{3\sqrt{3}}$$. I thought about using the identity
a³-b³ = (a-b)(a²+2ab+b²) as an alternative method with a=sin(x)+$$\frac{1}{\sqrt{3}}$$ and b=sin(x) . I find this :

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

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

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

 

[kaliprasad]

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
$\cos^3 x - \sin ^3 x = (\cos\,x - \sin\,x)^3 + 3 (\cos\,x - \sin\,x)(\cos\,x \sin\,x)$
$= \frac{1}{3\sqrt{3}} + 3 \frac{1}{\sqrt{3}} \frac{1}{3}$
$=\frac{1+3}{3\sqrt{3}}$
$=\frac{4}{3\sqrt{3}}$