Factor 2sin^2 x + 3cos x – 3 = 0

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In summary, the conversation is about solving a trigonometry equation by factoring and using the fundamental identity of circular trigonometry. The equation is solved by finding the solutions for cos x=1 and cos x=1/2 in the given interval of negative pi to pi. The solutions are 0 for cos x=1 and pi/3 and 5pi/3 for cos x=1/2.
  • #1
Liquidice_69
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im trying to solve this trig equation and the book says to factor it i think but its not clear, could anyone help. thxs

2sin^2 x + 3cos x – 3 = 0
 
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  • #2
[tex] sin^2x = 1 - cos^2x [/tex] is all you need. From there its a quadratic.
 
  • #3
but how do you get the cos^2
 
  • #4
Whozum gave you where that cos^2 comes from.U'd be using the fundamental identity of circular trigonometry

[tex] \sin^{2}x+\cos^{2}x\equiv 1 \ , \forall \ x\in\mathbb{R}[/tex]

Daniel.
 
  • #5
ok here's what i got so far

[tex] 2sin^2x+3cos x-3 = 0 [/tex]
[tex] 2(1-cos^2x)+3cos x-3 = 0 [/tex]
[tex] -2cos^2x+3cos x-1 = 0 [/tex]
[tex] 2cos^2x-3cos x+1 = 0 [/tex]

and i know how to factor
[tex]2x^2-3x+1=0 [/tex]
[tex](2x-1)(x-1)=0 [/tex]
[tex]x=1, x=.5 [/tex]

but how do i solve with the cos
 
  • #6
Well u get the equations

[tex] a) \ \cos x=1 [/tex]

[tex] b) \ \cos x=\frac{1}{2} [/tex]

Fix a domain where u want to solve these equations.

Daniel.
 
  • #7
ok, i want to solve it for negative pi to pi
 
  • #8
Well,then i can tell u that the fist equation has only solution if the interval is [itex] \left(-\pi,\pi\right) [/itex],and the second has 2.

Daniel.
 
  • #9
i don't get it, do u have to subtract pi from the second equasion
 
  • #10
[tex]\cos x=\frac{1}{2} [/tex] has the solution

[tex] x=\pm \frac{\pi}{3} [/tex] in the interval u were looking for the solution.

Daniel.
 
  • #11
ok i get it the answers are

for the cos x=1 0
for the cos x=.5 pi/3 and 5 pi/3
 
Last edited:

1. What is the equation "Factor 2sin^2 x + 3cos x – 3 = 0" used for?

The equation "Factor 2sin^2 x + 3cos x – 3 = 0" is used to solve for the values of x that satisfy the equation. It is commonly used in trigonometry and calculus problems.

2. How do I solve the equation "Factor 2sin^2 x + 3cos x – 3 = 0"?

To solve the equation "Factor 2sin^2 x + 3cos x – 3 = 0", you can use various algebraic and trigonometric identities and techniques such as factoring, substitution, and the quadratic formula. It is important to also consider the domain of the equation and make sure to check for extraneous solutions.

3. Can the equation "Factor 2sin^2 x + 3cos x – 3 = 0" have multiple solutions?

Yes, the equation "Factor 2sin^2 x + 3cos x – 3 = 0" can have multiple solutions since it is a quadratic equation in terms of sin x. This means that there can be two values of x that satisfy the equation, depending on the values of the coefficients.

4. What is the domain of the equation "Factor 2sin^2 x + 3cos x – 3 = 0"?

The domain of the equation "Factor 2sin^2 x + 3cos x – 3 = 0" is all real numbers since sin x and cos x can take on any value in the real number system. However, when solving the equation, it is important to consider any restrictions or limitations on x that may arise from the problem.

5. What does the graph of the equation "Factor 2sin^2 x + 3cos x – 3 = 0" look like?

The graph of the equation "Factor 2sin^2 x + 3cos x – 3 = 0" is a sinusoidal curve with a period of 2π and an amplitude of √(2/3). The x-intercepts of the graph represent the solutions to the equation, and the y-intercept is at y = -3. The graph can also be shifted vertically or horizontally depending on the values of the coefficients.

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