# Find the value of k:

• June 17th 2009, 10:07 AM
great_math
Find the value of k:
Find the values of $k$ for which the equation

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

is true
• June 17th 2009, 10:56 AM
running-gag
Quote:

Originally Posted by great_math
Find the values of $k$ for which the equation

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

is true

Hi

You can first get a quadratic in $\sin x$ by substituting $\cos^2 x - 1$ with $-\sin^2 x$
• June 18th 2009, 02:52 AM
Amer
Quote:

Originally Posted by great_math
Find the values of $k$ for which the equation

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

is true

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

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

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

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

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

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

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

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

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

so

$-\infty < k < \infty$ but I am not sure
• June 18th 2009, 06:13 AM
Amer
Quote:

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

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

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

or

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

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

let u=sinx

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

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

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

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

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

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

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

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

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

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

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

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

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

$0 \leq k \leq 4$

hmm (Itwasntme)
• June 18th 2009, 06:20 AM
Soroban
Hello, great_math!

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

Quote:

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

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

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

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

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

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

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

Good luck!

• June 19th 2009, 05:22 PM
pankaj
Putting $t=\sin\ x$.What we have is the quadratic

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

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

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

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

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

CASE 1

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

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

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

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

$|k-2|>1$

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

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

CASE 2

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

(2) $f(1)>0$

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

$f(-1)=2>0$

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

$\therefore |k-2|<1$

Thus $-1.

$k\in (1,3)$

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

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

On solving for $k$ we get $-2

Taking intersection of solutions of (1),(2) and (3),we get $1

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

$
\therefore k\in R
$