Solving Trigonometric Equation with Cosine and Sine Identities

[James889]

Hi,

I would like som help on this one:

Give all the real soulutions to the equation
$$\frac{7}{4}-2sin(x) - cos^2(x) = 0$$

i tried the common identities, like replacing $$cos^2(x)$$ with $$1-sin^2(x)$$

any ideas?

 

[Mentallic]

Well you picked the right identity. Why did you stop? :tongue:
You'll have a quadratic in sinx which you need to solve, and remember that the range of sinx is between -1 and 1 so scrap any solutions outside this range as they're not real.

 

[tiny-tim]

Hi James889!

i tried the common identities, like replacing $$cos^2(x)$$ with $$1-sin^2(x)$$

Good!

and now put sinx = y, and solve. ​

 

[James889]

Hi James889!


Good!

and now put sinx = y, and solve. ​

Thank you Flounder :),

So something along the lines of:
$$\frac{7}{4} -2sin(x)-cos^2(x) = 0$$

$$\frac{7}{4} -2sin(x)-1-sin^2(x) = 0$$

$$\frac{3}{4}-2sin(x)-sin^2(x)=0$$

substitute x = sin x

$$x^2 -2x -\frac{3}{4} = 0$$

$$-\frac{1}{2} \pm \sqrt{\frac{4+4*\frac{3}{4}}{2}}$$

 

[tiny-tim]

not fishy enough

nooo, not fishy enough

i] you got a minus sign wrong

ii] you still need to convert back

iii] using the same letter to mean two different things ("substitute x = sin x") is a really mammalian thing to do.

 

[James889]

nooo, not fishy enough

i] you got a minus sign wrong

ii] you still need to convert back

iii] using the same letter to mean two different things ("substitute x = sin x") is a really mammalian thing to do.

Hm, ye i think i messed up the quadratic formula i think it should be like this $$-x^2 -2x +\frac{3}{4} = 0$$

Giving me
$$\frac{1}{2} \pm \sqrt{\frac{(-2)^2+4*\frac{3}{4}}{2}}$$

hm, convert back?

 

[tiny-tim]

Hm, ye i think i messed up the quadratic formula i think it should be like this $$-x^2 -2x +\frac{3}{4} = 0$$

Giving me
$$\frac{1}{2} \pm \sqrt{\frac{(-2)^2+4*\frac{3}{4}}{2}}$$

No, almost completely wrong.

hm, convert back?

Yes, that'll be sinx, not x.

 

[James889]

Tim, tim, tim you're breaking my balls here...

one more try...deep breath

$$\frac{2}{2}\pm\sqrt{\bigg(\frac{-2}{2}\bigg)^2 -3/4}$$

So $$x = -0.5$$ but $$sin^{-1}(-0.5) = -30$$ that in the original equation gives 2
but $$sin^{-1}(0.5) = 30$$ gives the correct answer of 0.

 

[tiny-tim]

$$\frac{2}{2}\pm\sqrt{\bigg(\frac{-2}{2}\bigg)^2 -3/4}$$

Mysteriously, that's the right answer, but I don't know how you got it from -x² - 2x + 3/4 = 0

Nor how you get -0.5 from it

 

[Mentallic]

Obviously the quadratic formula method isn't working too well...

How about this?

$$-x^2 -2x +\frac{3}{4} = 0$$

$$4x^2+8x-3=0$$ (by multiplying through by -4)

Now, see if you can factorize this, because it can be factorized neatly.

 

[James889]

Obviously the quadratic formula method isn't working too well...

How about this?

$$-x^2 -2x +\frac{3}{4} = 0$$

$$4x^2+8x-3=0$$ (by multiplying through by -4)

Now, see if you can factorize this, because it can be factorized neatly.

$$(-2x+3)(2x-1)$$

But i don't get it, what was wrong with using the formula?

 

[tiny-tim]

$$(-2x+3)(2x-1)$$

But i don't get it, what was wrong with using the formula?

uhh? (-2x+3)(2x-1) = -4x² + 8x - 3

(and your formula started with b/2a instead of -b/2a)

 

[Mentallic]

As tiny-tim has said, you didn't factorize correctly. Try again and check your solution by expanding!

You want:

$$4x^2+8x-3$$

and you have:

$$(-2x+3)(2x-1)=-4x^2+8x-3$$

 

[James889]

Hm,

Is there some sort of trick when factorizing these kinds of expressions ?

i have only gone by trial and error so far, this could take a while :/

 

[tiny-tim]

I wouldn't bother to try to factor a quadratic equation unless the result is blindingly obvious …

it's quicker and safer to use the (-b ± √etc) formula.

 

[James889]

I wouldn't bother to try to factor a quadratic equation unless the result is blindingly obvious …

it's quicker and safer to use the (-b ± √etc) formula.

I tried but it turned out fishy ;)

uhh? (-2x+3)(2x-1) = -4x² + 8x - 3

(and your formula started with b/2a instead of -b/2a)

yes, i had $$-x^2 -2x +\frac{3}{4} = 0$$

-b in this case would be $$\frac{-(-2)}{2}$$

 

[tiny-tim]

I tried but it turned out fishy ;)

uhhh? what's wrong with being fishy?

yes, i had $$-x^2 -2x +\frac{3}{4} = 0$$

-b in this case would be $$\frac{-(-2)}{2}$$

Yes, you've been very messy , but you've got the right result at last: 1 ± 1/2.

ok, so if sinx = 1 ± 1/2, x = … ?​

 

[James889]

$$x = 30^{\circ}$$

Is happy tim ? :)

 

[tiny-tim]

Is happy tim ? :)

i'm always happy!

$$x = 30^{\circ}$$

Give all the real soulutions to the equation
$$\frac{7}{4}-2sin(x) - cos^2(x) = 0$$

mmm … so all the real solutions are … ?

 

[James889]

Oh, dang i forgot about that...
$$\frac{\pi}{6} +n*2\pi$$

 

[tiny-tim]

Oh, dang i forgot about that...
$$30^{\circ} +n*2\pi$$

oh come on …

you can't mix degrees and radians, can you?

and anyway, 30º isn't the only solution ≤ 360º. ​

 

[James889]

oh come on …

you can't mix degrees and radians, can you?

and anyway, 30º isn't the only solution ≤ 360º. ​

Heh, i knew that:uhh: i was just, um testing you

of course, silly me, we have $$sin(\pi-\pi/6)$$ as a solution aswell.

 

[tiny-tim]

Heh, i knew that:uhh: i was just, um testing you

of course, silly me, we have $$sin(\pi-\pi/6)$$ as a solution aswell.

ok, well i assume you're still testing me, so i think that would be …

5π/6 ​

 

[Mentallic]

Congrats tiny-tim, I think you passed the test!

I'm just being a little picky here: when you say another solution is $sin(5\pi/6)$ remember that the original equation was $-sin^2x-2sinx+3/4=0$ and the solutions to this equation are x=... so don't say a solution is $sin(5\pi/ 6)$ but rather just $5\pi /6$

Else if you were to test your solution, you'd be solving this: $$-sin^2(sin(5\pi/ 6))-2sin(sin(5\pi/ 6))+3/4$$ which does not equal 0 thus it's not a solution.


Ok so now you have the two solutions for $0\leq x\leq 2\pi$. $x=\pi/6$ and $x=5\pi/6$. Can you express all of the multiple values of these 2 solutions in just 1 equation? (I'm assuming that's how you're expected to show the values)

e.g.

$x=\pi /6 +2\pi n$
and
$x=5\pi /6 +2\pi n$
n being all integers.

Can you combine these 2 equations into 1?