determine the general solution of 3cossquaredx -5sinx=1

Question

What is the general solution and how do you find it? 

Answer 1

Let's start by using a trigonometric identity (sin²x + cos²x = 1) to replace that cos²x in the original equation. First, solve the trigonometric identity for cos²x:

 

cos²x = 1 - sin²x

 

Now replace cos²x in the original equation and move everything onto one side:

 

3(1 - sin²x) - 5sinx = 1

3 - 3sin²x - 5sinx = 1

3 - 3sin²x = 5sinx + 1

3 = 3sin²x + 5sinx + 1

0 = 3sin²x + 5sinx - 2

 

Now, you've got an unusual-looking quadratic equation. Just so you can picture it more clearly, let's replace sinx with y:

 

0 = 3y² + 5y - 2

 

This factors to:

 

0 = (3y - 1)(y + 2)

 

Solving for y, you get:

 

3y - 1 = 0, so y = 1/3

y + 2 = 0, so y = -2

 

Let's put sinx back into the equation instead of y:

 

sinx = 1/3 or -2

 

The values of the sine function are never lower than -1 or higher than 1, so the -2 answer is erroneous. We'll ignore it. That means that sinx = 1/3. Now, solve for x:

 

x = sin^-1(1/3) = 19.5º

 

This is one value for x that satisfies the equation. However, the problem asks for a general solution, meaning you must find all values of x that satisfy the equation. Since sine is a periodic function, repeating every 360º, we'll find a solution that works every 360º starting at 19.5º. You can express that like so:

 

x = 19.5º + 360ºn, where n is any integer

 

This means that your solutions include 19.5º, 379.5º (when n=1), 739.5º (when n=2), etc.