# Solving Quadratic Trig Equation

• Veronica_Oles
In summary: For sin x = 0 they did x = 2π-0 = 2πCan you conclude in an analogous way for ##sinx + 1 = 0##?For sin x = 0 they did x = 2π-0 = 2π ...If they actually stated that in their solution, then that's in error and what you say after that is correct.
Veronica_Oles

## Homework Statement

Solve sin2x + sinx =0

## The Attempt at a Solution

I first factor and get
sinx(sinx + 1) = 0

sinx = 0
x = sin-1(0)
x= 0
x= π-0 = π

sinx = -1
x= sin-1(1)
x= π/2
x = π + π/2 = 3π/2

x = { 0, π, 3π/2 }

However when I look at the solution this is correct but they also added 2π and I'm not sure why this is?
It seems like for sinx = 0 they did x = 2π-0 = 2π why is this? I thought that if it's zero you obtain your other value through the sin quadrant.

Veronica_Oles said:

## Homework Statement

Solve sin2x + sinx =0

## The Attempt at a Solution

I first factor and get
sinx(sinx + 1) = 0

sinx = 0
x = sin-1(0)
x= 0
x= π-0 = π

sinx = -1
x= sin-1(1)
x= π/2
x = π + π/2 = 3π/2

x = { 0, π, 3π/2 }

However when I look at the solution this is correct but they also added 2π and I'm not sure why this is?
It seems like for sinx = 0 they did x = 2π-0 = 2π why is this? I thought that if it's zero you obtain your other value through the sin quadrant.

For ##sinx = 0, x = n\pi## because ##sinx## becomes zero for ##n = 0, 1, 2, \cdots##
Can you conclude in an analogous way for ##sinx + 1 = 0##?

Veronica_Oles said:
x = { 0, π, 3π/2 }

However when I look at the solution this is correct but they also added 2π and I'm not sure why this is?
It seems like for sinx = 0 they did x = 2π-0 = 2π why is this? I thought that if it's zero you obtain your other value through the sin quadrant.
If X is a solution, or X+2kπ is also solution where k is integer. So the solutions of the problem are 0+2kπ; π+2nπ; 3π/2+2mπ,(k, m, n are integers).

Possibly, the problem asked the solutions in the closed interval [0, 2π]. In this case, both 0 and 2π have to be included.

You wrote:
Veronica_Oles said:
It seems like for sin x = 0 they did x = 2π-0 = 2π ...
If they actually stated that in their solution, then that's in error and what you say after that is correct.

However, if it simply "seems" to you that they did this to get 2π, then I suspect that they did not use that reasoning, but merely used the fact that sin(x) has a period of 2π so that if x = 0 is a solution then x = 0 + 2π is also a solution.

Veronica_Oles said:

## Homework Statement

Solve sin2x + sinx =0

## The Attempt at a Solution

I first factor and get
sinx(sinx + 1) = 0

sinx = 0
x = sin-1(0)
x= 0
x= π-0 = π

sinx = -1
x= sin-1(1)

You mean ##x = \arcsin(-1)##, which doesn't give you ##\frac \pi 2##. It gives you ##x = -\frac \pi 2##. Your ##\frac{3\pi} 2## below is correct but your reasoning is incorrect.

x= π/2
x = π + π/2 = 3π/2

x = { 0, π, 3π/2 }

LCKurtz said:
You mean ##x = \arcsin(-1)##, which doesn't give you ##\frac \pi 2##. It gives you ##x = -\frac \pi 2##. Your ##\frac{3\pi} 2## below is correct but your reasoning is incorrect.
x = sin-1(-1)
x = -π/2
x = π-(-π/2)
x = 3π/2

Veronica_Oles said:
x = sin-1(-1)
x = -π/2
x = π-(-π/2)
x = 3π/2

That gets ##x=\frac \pi 2## alright, but if I wanted an angle in ##[0,2\pi]##, I would simply take ##-\frac \pi 2 + 2\pi##, which would give ##\frac {3\pi} 2##. I don't understand your thinking to calculate ##\pi - (-\frac \pi 2)##. On what basis would you calculate that way?

## 1. How do I solve a quadratic trigonometric equation?

To solve a quadratic trigonometric equation, you must first isolate the trigonometric term on one side of the equation. Then, use the appropriate trigonometric identities and properties to rewrite the equation in terms of a single trigonometric function. Finally, use algebraic techniques such as factoring or the quadratic formula to solve for the variable.

## 2. What are the common trigonometric identities used in solving quadratic trigonometric equations?

Some common trigonometric identities used in solving quadratic trigonometric equations include the Pythagorean identities (sin^2x + cos^2x = 1), the double angle identities (sin2x = 2sinx cosx, cos2x = cos^2x - sin^2x), and the sum and difference identities (sin(x ± y) = sinx cosy ± cosx siny, cos(x ± y) = cosx cosy ∓ sinx siny).

## 3. Can a quadratic trigonometric equation have more than one solution?

Yes, a quadratic trigonometric equation can have up to two solutions, or roots. This is because a quadratic equation can have two real solutions, or two complex solutions. However, in some cases, a quadratic trigonometric equation may only have one real solution or no real solutions.

## 4. What are some tips for solving difficult quadratic trigonometric equations?

Some tips for solving difficult quadratic trigonometric equations include factoring, using the quadratic formula, and checking for extraneous solutions. It is also helpful to have a strong understanding of trigonometric identities and properties.

## 5. How can I check my solutions for a quadratic trigonometric equation?

To check your solutions for a quadratic trigonometric equation, you can substitute the values back into the original equation and see if they satisfy the equation. You can also use a graphing calculator to graph the equation and see if the solutions intersect the graph at the correct points.

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