Logarithm Question (2)

$\log_{\sqrt{2} \sin x} (1-\sin x) = 2 $

Question

Logarithm Question (2) 

$\log_{\sqrt{2} \sin x} (1-\sin x) = 2 $

mathslover · Answer

@ParthKohli @myininaya @jim_thompson5910

tkhunny · Answer

What?  Is that $\sqrt{2}\sin(x)$ the Base of the Logarithm?

mathslover · Answer

Yes.

anonymous · Answer

ITs the base

mathslover · Answer

There are 2 methods that point towards a single result :

$\bf{\bullet}$ First Method :

$\log_{\sqrt{2} \sin x} (1-\sin x) = \log_{\sqrt{2} \sin x} ( \sqrt{2} \sin x)^2 $ 

Therefore, $1-\sin x  = 2\sin^2 x$

tkhunny · Answer

Weird.  Don't think I've ever seen that.

You seem to have transformed it to the equivalent exponential form.  That's good.

parthkohli · Answer

Whoops, sorry.

parthkohli · Answer

First, keep the conditions for a logarithm to be defined in your mind.  Solve it only then.

mathslover · Answer

$\bf{\bullet}$ Second Method :

$\cfrac{1}{2} \log_{\sqrt{2}\sin x} (1-\sin x) = 1 \

\implies  \log_{{\left(\sqrt{2} \sin x\right) }^2} (1-\sin x) = 1 \
\implies  \log_{2\sin^2 x } (1-\sin x) = 1 \
\implies  1 -\sin x = 2 \sin^2 x $

parthkohli · Answer

Yes, solve the quadratic.$$\sqrt{2}\sin(x) > 0 \implies \sin(x) > 0,\sqrt{2}\sin(x) \ne 1$$$$1  - \sin(x) > 0 \implies \sin(x) < 1 \checkmark  $$

tkhunny · Answer

Time!!  I want to think about Domain issues.  I always want to think about that.

The definition of a base requires that it not be one (1).  Your budy $\sqrt{2}\sin(x)$ is often one (1).  What's your plan for resolving that?

mathslover · Answer

We have an equation now : $1-\sin x = 2\sin ^2 x$ 

Again, this can be solved in 2 ways :

$\bf{\bullet}$ First Method : 

$\sin x = y$ ; (Say) 

So, $1 - y = 2y^2 \

\implies 2y^2 + y - 1=0 $ 

Using Quadratic Equation formula : 

$y = \cfrac{-1 \pm \sqrt{9}}{4} = \cfrac{-1 \pm 3}{4} $ 

$y = \cfrac{1}{2}$ or $y = -1$

parthkohli · Answer

Why not just factor?  :P$$\sin(x) > 0$$

mathslover · Answer

Well, if I take it directly @tkhunny :

From log properties : 

1 - sinx > 0 

$\implies x < \cfrac{\pi}{2}$ 

As, $\sqrt{2} \sin x \ne 1$ 

Therefore, $x \ne \cfrac{\pi}{4}$

mathslover · Answer

And from the first method we got y = 1/2 or y = -1

That is $\sin x = 1/2$ or $\sin x = -1$

parthkohli · Answer

You don't need to do those.  You have the values for $\sin(x)$, so there is no need to find the angles for the conditions.

parthkohli · Answer

Just eliminate $\sin(x) = -1$ as $\sin(x) >0$.

tkhunny · Answer

Where did we get $\sin(x) > 0$?  That makes no sense.  Did I miss an assumption along the way?

mathslover · Answer

@tkhunny , the base.

mathslover · Answer

$\sqrt{2} \sin x > 0$

mathslover · Answer

implies that : sin x > 0

mathslover · Answer

So, sin x can not be equal to 1.

Thus, we have : sin x = 1/2

x = pi/6

mathslover · Answer

For the second method to solve 1 - sin x = $2\sin^2 x$ 

We can do it like this also :

$1-2\sin^2 x = \sin x$

$\cos (2x) = \sin x$

mathslover · Answer

My main question, was just a very little equation (above) : cos(2x) = sin x

How to conclude the answer from here?

anonymous · Answer

double angle formula

anonymous · Answer

cos(2x) = cos^2(x) - sin^2(x)
and cos^2(x) = 1 - sin^2(x)

mathslover · Answer

I meant to say that, how to get the value of "x" from

cos(2x) = sin(x)

tkhunny · Answer

I was arguing that with myself for a little while.  I yield.

anonymous · Answer

everything now is in terms of sin(x). Solve it

anonymous · Answer

cos(2x) = sin(x)
(1 - sin^2(x)) - sin^2(x) = sin(x)

mathslover · Answer

Yes, we can solve it by 1 - sin x = 2sin^2 x

but, I wanted to know that is there any way that I can solve it directly from here :

cos(2x) = sin (x) ?

mathslover · Answer

without  simplifying it again into the original form (1-sin x = 2sin^2 x)

anonymous · Answer

well i don't know what you mean by solving directly. As far as I know, there is now way you can solve it without using trig identity. You could perhaps take the arccosine for both sides, but then you would need to know the properties of composition of trig inverse functions. This seems to make things rather more complicated

anonymous · Answer

there is *no* way

mathslover · Answer

Oh... thanks @sourwing ... I will prefer not to make it more complicated, and as I have not learnt trig. inverse functions, so, I will not try that either. :)

mathslover · Answer

Thank you @tkhunny @sourwing @ParthKohli .

anonymous · Answer

and there is actually infinitely many solutions to original equation ^^ https://www.wolframalpha.com/input/?i=log_%7Bsqrt%282%29+sin%28x%29%7D+%281+-+sin%28x%29%29+%3D+2

parthkohli · Answer

You can try to see the unit circle.