# Solve Simple Trig Problem: cosx - cosx*sinx = 1/3

• Mddrill
In summary, the conversation discusses the difficulty in evaluating the equation cosx - cosx*sinx = 1/3. The individual attempted to solve it using trigonometric identities but was unsuccessful. They then referenced a website that provided a numerical solution but not a simplified trigonometric form. Maple 11 is mentioned to have obtained an "analytic" solution, but it is complex and involves finding roots of a polynomial. The final summary is that the equation is simple to solve numerically but not analytically.
Mddrill

## Homework Statement

I feel like I this problem shouldn't be that hard, but I can't figure out how to evaluate this equation

cosx - cosx*sinx = 1/3

## Homework Equations

I don't know what to use for this, none of the trig identities seem to help

## The Attempt at a Solution

I tried substituting ## sqrt(1-cos^2x )## for sinx , but I just ended up with something way more complicated. I know this shouldn't be this hard because I'm not even in a trig class, I am taking statics. It wouldn't make sense for them to give me a problem that requires more time with the trig than with the statics.

Thank You

Mddrill said:

## Homework Statement

I feel like I this problem shouldn't be that hard, but I can't figure out how to evaluate this equation

cosx - cosx*sinx = 1/3

## Homework Equations

I don't know what to use for this, none of the trig identities seem to help

## The Attempt at a Solution

I tried substituting ## sqrt(1-cos^2x )## for sinx , but I just ended up with something way more complicated. I know this shouldn't be this hard because I'm not even in a trig class, I am taking statics. It wouldn't make sense for them to give me a problem that requires more time with the trig than with the statics.

Thank You

WolframAlpha.com shows solutions to this, but it doesn't seem to show a simplified trig form...

Mddrill
Mddrill said:

## Homework Statement

I feel like I this problem shouldn't be that hard, but I can't figure out how to evaluate this equation

cosx - cosx*sinx = 1/3

## Homework Equations

I don't know what to use for this, none of the trig identities seem to help

## The Attempt at a Solution

I tried substituting ## sqrt(1-cos^2x )## for sinx , but I just ended up with something way more complicated. I know this shouldn't be this hard because I'm not even in a trig class, I am taking statics. It wouldn't make sense for them to give me a problem that requires more time with the trig than with the statics.

Thank You

This is simple to solve numerically, but not at all easy to solve "analytically" (i.e., using formulas). For what it's worth, Maple 11 obtains the following "analytic" solution:
x =
arctan((-1+3*(1/6*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+1/6*((-(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)*(162+6*681^(1/2))^(2/3)-12*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+36*(162+6*681^(1/2))^(1/3))/(162+6*681^(1/2))^(1/3)/(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2))^(1/2))^3)/(1/6*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+1/6*((-(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)*(162+6*681^(1/2))^(2/3)-12*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+36*(162+6*681^(1/2))^(1/3))/(162+6*681^(1/2))^(1/3)/(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2))^(1/2)))
Basically, this has the form
$$x = \arctan\left( \frac{-1 + 3z^3}{z} \right),$$
where ##z## is a root of the equation ##9z^4 -6z + 1 = 0##. Numerically, ##x \doteq 0.6284918352##. Of course, there are other solutions outside the region ##0 < x < \pi##.

Mddrill

## 1. What is the first step in solving this trigonometry problem?

The first step in solving this problem is to simplify the left side of the equation by factoring out a common factor of cosx. This will result in cosx(1-sinx) = 1/3.

## 2. How can I solve for sinx in this equation?

To solve for sinx, you can divide both sides of the equation by cosx, resulting in 1-sinx = 1/3cosx. Then, you can subtract 1 from both sides and use the trigonometric identity sin^2x + cos^2x = 1 to simplify the equation and solve for sinx.

## 3. Can this equation have multiple solutions?

Yes, this equation can have multiple solutions. In fact, it has infinitely many solutions since the sine function is periodic and repeats itself every 2π radians (or 360 degrees).

## 4. How can I check my solution for this equation?

You can check your solution by plugging it back into the original equation and seeing if it satisfies the equation. You can also use a calculator to evaluate the left and right sides of the equation and see if they are equal.

## 5. Are there any special cases to consider when solving this trigonometry problem?

Yes, there are a few special cases to consider. For example, if the equation is undefined when dividing by cosx, then the solution would not be valid. Additionally, you may need to consider the domain and range of the sine function when solving for sinx.

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