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

• MHB
• sp3
In summary, the conversation discusses solving the equation cos(x) = sin(x) + 1/sqrt(3) in order to find the value of cos(x)3 - sin3(x). The process involves using the identity sin^2(x) + cos^2(x) = 1 and the quadratic formula to find the two solutions for sin(x). Then, using the identity a^3 - b^3 = (a-b)(a^2 + 2ab + b^2) and substituting in the values for sin(x) and cos(x), the final solution for cos(x)3 - sin3(x) is found to be 4/3sqrt(3). The conversation also mentions the helpfulness of
sp3
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!

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.

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

## 1. What is a trigonometric equation?

A trigonometric equation is an equation that involves trigonometric functions, such as sine, cosine, and tangent, and an unknown variable. The goal is to solve for the value of the variable that makes the equation true.

## 2. How do you solve a trigonometric equation?

To solve a trigonometric equation, you can use algebraic manipulation and trigonometric identities to simplify the equation and isolate the variable. You can also use a calculator or graphing software to find the solutions.

## 3. How do you solve the equation cos(x)=sin(x) + 1/√3?

To solve this equation, you can use the Pythagorean identity, cos^2(x) + sin^2(x) = 1, to rewrite the equation as cos(x) = √(1-sin^2(x)). Then, you can substitute this expression for cos(x) into the original equation and solve for sin(x). Once you have the value of sin(x), you can use inverse trigonometric functions to find the solutions for x.

## 4. What are the solutions to the equation cos(x)=sin(x) + 1/√3?

The solutions to this equation are x = π/6 + 2πn and x = 5π/6 + 2πn, where n is any integer. These solutions can be found by using inverse trigonometric functions to solve for sin(x) and then substituting the values into the original equation.

## 5. How do you check if a solution is valid for the equation cos(x)=sin(x) + 1/√3?

To check if a solution is valid, you can substitute the value of x into the original equation and see if it makes the equation true. You can also graph the equation and the solution on a graphing calculator or software and see if they intersect at the given x value.

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