Solving y = sec^2x - 2sin x: Tips & Strategies

In summary: Similarly cos3(x) means to cube the cosine:cos3(x) = (cos(x))3In summary, the conversation is about solving the equation y = sec^2 x - 2sin x and the confusion about using exponents with trigonometric functions. The equation can be simplified by using the identity sec^2 x = 1/cos^2 x and the value of cos(pi/4) is needed to evaluate the equation at x = pi/4. The concept of using exponents with trigonometric functions is also clarified.
  • #1
abm7
1
0
Post was misplaced in a technical forum section, so is missing the homework template
I need to solve y = sec^2 x - 2sin x, I'm fine with the 2sin x, but for the sec^2 x...

I can't seem to understand how to do it... I can get it to...

sin^2 x + cos^2 x / cos^2 x or 1 / cos^2x

but I don't understand how to solve it when the square comes right after the cos/sin etc.
 
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  • #2
abm7 said:
I need to solve y = sec^2 x - 2sin x, I'm fine with the 2sin x, but for the sec^2 x...
What do you mean by "solving" this equation?
To solve an equation that involves a single variable, you isolate the variable on one side of the equation, and everything else on the other side.
When you solve an equation that involves two variables, you isolate the variable you're solving for on one side, with everything else on the other.##\sec x = \frac{1}{\cos x}##, so ##\sec^2 x = \frac{1}{\cos^2 x}##. What is ##\cos(\pi/4)##?
Your equation is already solved for y, and I don't think it's possible to solve for x in that equation.

Edit: Ah, now I understand. You want to evaluate ##\sec^2 x - 2\sin x## at ##x = \pi/4##.
This information should be in the body of the post, not just in the thread title.
abm7 said:
I can't seem to understand how to do it... I can get it to...

sin^2 x + cos^2 x / cos^2 x or 1 / cos^2x

but I don't understand how to solve it when the square comes right after the cos/sin etc.
 
Last edited:
  • #3
Now I am confused! abm7, do you want to solve [itex]y= sec^2(x)- 2sin(x)[/itex] for x, in terms of y, or do you want to evaluate it at [itex]x= \pi/4[/itex]?

"Solving" the equation will be hard- it reduces to a cubic equation for sin(x).
 
  • #4
abm7 said:
I need to solve y = sec^2 x - 2sin x, I'm fine with the 2sin x, but for the sec^2 x...

I can't seem to understand how to do it... I can get it to...

sin^2 x + cos^2 x / cos^2 x or 1 / cos^2x

but I don't understand how to solve it when the square comes right after the cos/sin etc.
You need to enclose the entire numerator in parentheses as follows/
(sin^2 x + cos^2 x) / cos^2 x​

I get the sense that you don';t understand what is meant by using the exponent with the trig functions such as
sin2(x)​

All it means is to take sin(x) and square it:
sin2(x) = (sin(x))2
 

Related to Solving y = sec^2x - 2sin x: Tips & Strategies

1. What is the first step in solving y = sec^2x - 2sin x?

The first step is to rewrite the equation in terms of a single trigonometric function. In this case, we can rewrite sec^2x as 1/cos^2x. So the equation becomes y = 1/cos^2x - 2sin x.

2. How do you simplify the equation y = 1/cos^2x - 2sin x?

To simplify the equation, we can use the identity sin^2x + cos^2x = 1. Substituting this into the equation, we get y = 1/cos^2x - 2(sin^2x + cos^2x). This simplifies to y = -cos^2x - 2sin^2x + 2.

3. What are some strategies for solving trigonometric equations?

One strategy is to use trigonometric identities to rewrite the equation in terms of a single trigonometric function. Another strategy is to use substitution, where you replace one trigonometric function with another. Additionally, graphing the equation can help identify solutions.

4. How do you solve for x in the equation y = -cos^2x - 2sin^2x + 2?

To solve for x, we need to isolate the trigonometric function on one side of the equation. We can rearrange the equation to get -cos^2x - 2sin^2x = y - 2. Then, we can use the Pythagorean identity, sin^2x + cos^2x = 1, to get -1 - cos^2x = y - 2. From here, we can solve for cos^2x and then take the square root to find the values of x.

5. What are the possible solutions for x in the equation y = -cos^2x - 2sin^2x + 2?

The possible solutions for x depend on the value of y. If y is a real number between -1 and 1, there will be two solutions for x. If y is less than -1 or greater than 1, there will be no real solutions for x. This is because the cosine function is only defined between -1 and 1. If y is equal to -1 or 1, there will be infinitely many solutions for x. These values are known as the asymptotes of the graph of y = -cos^2x - 2sin^2x + 2.

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