Solving a trigonometric equation

[coooool222]

Solve the equation (Enter your answers as a comma-separated list. Use as an integer constant. Enter your response in radians)
[math]2 cos^2(x) + 5 sin(x) = 4[/math]​

I have no idea what to do whatsoever. I know you have to use the quadratic formula i did that.

x [math]x = -5 + \sqrt{57} / 4[/math][math]x = -5 - \sqrt{57} / 4[/math]
I have the two x products now what do I do.

 

[Dr.Peterson]

Solve the equation (Enter your answers as a comma-separated list. Use as an integer constant. Enter your response in radians)
[math]2 cos^2(x) + 5 sin(x) = 4[/math]​

I have no idea what to do whatsoever. I know you have to use the quadratic formula i did that.

x [math]x = -5 + \sqrt{57} / 4[/math][math]x = -5 - \sqrt{57} / 4[/math]
I have the two x products now what do I do.

How did you get those values? Please show your work, so we can correct it.

Did you first rewrite the equation in terms of sine only? If so, you made some other mistake. (And solving a quadratic will not give you x itself, but sin(x), or something of the sort.)

 

[coooool222]

why sin x

 

[Dr.Peterson]

why sin x

There may be other ways, but you have to get it in terms of some one trig function.

Once you show your work, as we ask, we can point out errors more specifically.

 

[coooool222]

A =2 b =5 C = -4

X = -5 +- SQRT 5^2-4(2)(-4) / 2(2)

[math]x = -5 + \sqrt{57} / 4[/math]
[math]x = -5 - \sqrt{57} / 4[/math]

 

[coooool222]

idid it

 

[BigBeachBanana]

idid it

You can't do that quite yet. First rewrite [imath]cos^2(x) = 1 - sin^2(x)[/imath]. Can you simplify and go from there? Please post back your work with all steps?

 

[Dr.Peterson]

A =2 b =5 C = -4

X = -5 +- SQRT 5^2-4(2)(-4) / 2(2)

[math]x = -5 + \sqrt{57} / 4[/math]
[math]x = -5 - \sqrt{57} / 4[/math]

This would be a solution if the problem were [imath]2x^2+5x-4=0[/imath]. But it isn't. It's [imath]2(\cos x)^2+5(\sin x)-4=0[/imath].

Your x's don't even replace the same quantity in the actual equation. Please do what BBB (and I earlier) said to do, and get everything in terms of one trig function, which you can then replace with a variable (not x, but maybe u).

Actually, your work is wrong even for the equation you thought it was, because you didn't parenthesize the entire numerator. It would be [math]x = (-5 \pm \sqrt{57}) / 4 = \frac{-5 \pm \sqrt{57}} {4}[/math] What you wrote is [math]x = -5 \pm \sqrt{57} / 4 = -5 \pm\frac{\sqrt{57}} {4}[/math]

 

[topsquark]

To expand a bit on what's being said above:
[imath]2 ~ cos^2(x) + 5 ~ sin(x) - 4 = 0[/imath]

[imath]2 ~ (1 - sin^2(x)) + 5 ~ sin(x) - 4 = 0[/imath]

[imath]2 - 2 ~ sin^2(x) + 5 ~ sin(x) - 4 = 0[/imath]

[imath]-2 ~ sin^2(x) + 5 ~ sin(x) - 2 = 0[/imath]

Now let y = sin(x)
[imath]-2y^2 + 5y - 2 = 0[/imath]

Solve for y. Then let [imath]x = sin^{-1}(y)[/imath].

You aren't going to be able to find a nice angle for x.

-Dan

 

[Dr.Peterson]

You aren't going to be able to find a nice angle for x.

Actually, the solution is a nice angle ... or rather, infinitely many nice angles. (The problem says "Use as an integer constant", which I suppose is supposed to have been something like "Use k as an integer constant".)