# Thread: Find the value of k:

1. ## Find the value of k:

Find the values of $\displaystyle k$ for which the equation

$\displaystyle \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1$

is true

2. Originally Posted by great_math
Find the values of $\displaystyle k$ for which the equation

$\displaystyle \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1$

is true
Hi

You can first get a quadratic in $\displaystyle \sin x$ by substituting $\displaystyle \cos^2 x - 1$ with $\displaystyle -\sin^2 x$

3. Originally Posted by great_math
Find the values of $\displaystyle k$ for which the equation

$\displaystyle \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1$

is true

$\displaystyle sin^2x - \mid k-2 \mid sinx = cos^2 x + \mid k-2 \mid -1$

$\displaystyle sin^2 x - \mid k-2 \mid sinx = -sin^2x + \mid k-2 \mid$

$\displaystyle 2sin^2x = (\mid k-2 \mid) (sinx +1 )$

$\displaystyle \frac{2sin^2x }{sinx +1 } = \mid k-2 \mid$

$\displaystyle k-2 = \frac{2sin^2x }{sinx +1 } \Rightarrow k=\frac{2sin^2x }{sinx +1 }+2$

$\displaystyle k-2 = \frac{-(2sin^2x) }{sinx +1 }\Rightarrow k=\frac{-(2sin^2x) }{sinx +1 }+2$

$\displaystyle x \ne \frac{3\pi}{2} + 2n\pi$ since if you do not put this condition the denominator will equal zero and k will be infinity so

$\displaystyle k=\frac{2sin^2x }{sinx +1 }+2 \Rightarrow 2\leq k < \infty$

$\displaystyle k=\frac{-(2sin^2x) }{sinx +1 }+2 \Rightarrow -\infty< k \leq 2$

so

$\displaystyle -\infty < k < \infty$ but I am not sure

4. Originally Posted by Amer
$\displaystyle sin^2x - \mid k-2 \mid sinx = cos^2 x + \mid k-2 \mid -1$

$\displaystyle sin^2 x - \mid k-2 \mid sinx = -sin^2x + \mid k-2 \mid$

$\displaystyle 2sin^2x = (\mid k-2 \mid) (sinx +1 )$

or

$\displaystyle 2sin^2x = \mid k-2 \mid (sin(x))+\mid k-2 \mid$

$\displaystyle 2sin^2x -\mid k-2 \mid (sin(x))-\mid k-2 \mid =0$

let u=sinx

$\displaystyle 2u^2 -\mid k-2 \mid (u) -\mid k-2 \mid =0$

$\displaystyle u=\frac{\mid k-2 \mid \mp \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}$

you know that $\displaystyle -1\leq sinx\leq 1$

$\displaystyle -1\leq \frac{\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}\leq 1$

$\displaystyle -1\leq \frac{\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}\leq 1$

$\displaystyle -4\leq (\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )})\leq 4$

$\displaystyle -4\leq (\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )})\leq 4$

$\displaystyle 0\leq [\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}]\times$$\displaystyle [\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}]\leq 16$

$\displaystyle 0\leq (\mid k-2 \mid)^2 - [(\mid k-2 \mid)^2 -8\mid k-2 \mid] \leq 16$

$\displaystyle 0 \leq 8 \mid k-2 \mid \leq 16$

$\displaystyle 0 \leq \mid k-2 \mid \leq 2$

$\displaystyle \mid k-2 \mid \leq 2$

$\displaystyle -2\leq k-2 \leq 2$

$\displaystyle 0 \leq k \leq 4$

hmm

5. Hello, great_math!

A puzzling problem . . . What is meant by "is true"?

Find the values of $\displaystyle k$ for which the equation: .$\displaystyle \sin^2\!x-|k-2|\sin x=\cos^2\!x+|k-2|-1$ .is true.
Replace $\displaystyle \cos^2\!x$ with $\displaystyle 1 - \sin^2\!x$:

. . $\displaystyle \sin^2\!x - |k-2|\sin x \;=\;1 - \sin^2\!x +|k-2| - 1$

. . $\displaystyle 2\sin^2\!x - |k-2|\sin x - |k-2| \;=\;0$

Quadratic Formula: . $\displaystyle \sin x \;=\;\frac{|k-2| \pm\sqrt{|k-2|^2 + 8|k-2|}}{4}$

Now it all depends on what they're asking for . . .

. . The discriminant must be nonnegative: .$\displaystyle |k-2|^2 + 8|k-2| \;\geq \;0$

. . Since $\displaystyle |\sin x| \:\leq \:1\!:\quad \left|\frac{|k-2| \pm\sqrt{|k-2|^2 + 8|k-2|}}{4}\right| \;\leq \;1$

Good luck!

6. Putting $\displaystyle t=\sin\ x$.What we have is the quadratic

$\displaystyle 2t^2-|k-2|t-|k-2|=0$ where $\displaystyle t\in[-1,1]$

Discriminant = $\displaystyle |k-2|^2+8|k-2|\geq 0$

Thus we will get real values of $\displaystyle t$ for all real values of $\displaystyle k$.

But we are required to ensure that these values must lie on the interval $\displaystyle [-1,1]$

Let $\displaystyle f(t)=2t^2-|k-2|t-|k-2|=0$

CASE 1

If exactly one root of $\displaystyle f(t)=0$ lies on (-1,1),then we must have

$\displaystyle f(-1)f(1)<0$

Now,$\displaystyle f(-1)=2>0$

$\displaystyle \therefore f(1)<0$,thus $\displaystyle 2-2|k-2|<0$

$\displaystyle |k-2|>1$

$\displaystyle k-2<-1$ or $\displaystyle k-2>1$

$\displaystyle k<1$ or $\displaystyle k>3$i.e.$\displaystyle k\in(-\infty,1)\cup(3,\infty)$

CASE 2

If both roots on the interval $\displaystyle (-1,1)$ then following three conditions must be satisfied:
(1)$\displaystyle f(-1)>0$

(2)$\displaystyle f(1)>0$

(3)$\displaystyle -1<\frac{\alpha+\beta}{2}<1$;$\displaystyle \alpha$ and $\displaystyle \beta$ being the rootsof the equation $\displaystyle f(t)=0$

$\displaystyle f(-1)=2>0$

$\displaystyle f(1)>0$ implies that $\displaystyle 2-2|k-2|>0$.

$\displaystyle \therefore |k-2|<1$

Thus $\displaystyle -1<k-2<1$.

$\displaystyle k\in (1,3)$

$\displaystyle -1<\frac{\alpha+\beta}{2}<1$ implies that $\displaystyle -1<\frac{|k-2|}{4}<1$

(Recall that if $\displaystyle \alpha$and $\displaystyle \beta$are roots of $\displaystyle ax^2+bx+c=0$ then $\displaystyle \alpha+\beta=-\frac{b}{a}$)

On solving for $\displaystyle k$ we get $\displaystyle -2<k<6$

Taking intersection of solutions of (1),(2) and (3),we get $\displaystyle 1<k<3$

Finally taking union of Case 1 and Case 2 and also checking for $\displaystyle k=1,3$, we arrive at the conclusion that the given equation is true for all real $\displaystyle k$.

$\displaystyle \therefore k\in R$ must be the answer.