Find real solutions to an equation

[diredragon]

Homework Statement

log(cosx)sinx = 4*log(sinx)cosx

Homework Equations

3. The Attempt at a Solution
i tried to solve it and uploaded my work but my last part reads
(cosx)^2 = sinx
(cosx)^(-2) = sinx
its weird that i don't see where i have made a mistake[/B]

 

[SammyS]

Homework Statement

log(cosx)sinx = 4*log(sinx)cosx

Homework Equations

3. The Attempt at a Solution [/B]
i tried to solve it and uploaded my work but my last part reads
(cosx)^2 = sinx
(cosx)^(-2) = sinx
its weird that i don't see where i have made a mistake

First of all:
Do you really mean (log(cos x))⋅sin x = 4⋅(log(sin x))⋅cos x ? (Mathematically speaking, that is what you wrote.)

Furthermore, it's difficult to say where you went wrong, without you giving more detail in your workings.

 

[diredragon]

What is in the brackets i meant to be the base.

 

[SammyS]

So, your initial problem must have had:

log_(cosx)(sinx) = 4*log_(sinx)(cosx) .​

 

[diredragon]

So, your initial problem must have had:

log_(cosx)(sinx) = 4*log_(sinx)(cosx) .​

yes and I am trying to show my work but the pic won't upload. Ill try to fix it

 

[diredragon]

http://s29.postimg.org/il7yopzx3/20151123_155834.jpg
Here it is.

 

[SammyS]

(cosx)^2 = sinx
(cosx)^(-2) = sinx
It's weird that i don't see where i have made a mistake

http://s29.postimg.org/il7yopzx3/20151123_155834.jpg
Here it is.

Sideways image.

I get the same result using change of base formula.

Rewrite that second result as:
1 = (sin x)(cos² x)

I think only the first one has real number solutions.

 

[Mark44]

Homework Statement

log(cosx)sinx = 4*log(sinx)cosx

As already noted, the above appears to be $\log(\cos x) \cdot \sin x = 4 \log(\sin x) \cdot \cos x$
As it turns out, you intended the log bases to be cos(x) and sin(x), respectively

To make you work understandable, use either BBCode subscripts or LaTeX to write these expressions. Below are the a couple of examples of each, showing how they are formed and how they appear.
BBCode: log[SUB]cos(x)[/SUB](sin(x)) - renders as log_cos(x)(sin(x) -- see https://www.physicsforums.com/help/bb-codes
LaTeX: ##\log_{\cos(x)}\sin(x)## - renders as $\log_{\cos(x)}\sin(x)$ -- see https://www.physicsforums.com/help/latexhelp/

Homework Equations

3. The Attempt at a Solution [/B]
i tried to solve it and uploaded my work but my last part reads
(cosx)^2 = sinx
(cosx)^(-2) = sinx
its weird that i don't see where i have made a mistake

 

[diredragon]

I get
-(sinx)^2 + 1 = sinx
And
1 = sinx - (sinx)^3 for the two cases
How do i now find the real solutions out of this?

 

[Samy_A]

I get
-(sinx)^2 + 1 = sinx
And
1 = sinx - (sinx)^3 for the two cases
How do i now find the real solutions out of this?

The first is a quadratic equation (in sin(x)).
The second one you better leave in the form @SammyS used, and try to see why he thought that it doesn't have a real solution.

 

[diredragon]

From the first quadratic equation i get sinx=(-1+-(5)^(1/2))/2 the only real solution is + one because log is not defined for negative numbers. Using the identity i get cosx to be the same. How now do i obtain the real solutions? Is this correct?

 

[Samy_A]

From the first quadratic equation i get sinx=(-1+-(5)^(1/2))/2 the only real solution is + one because log is not defined for negative numbers. Using the identity i get cosx to be the same. How now do i obtain the real solutions? Is this correct?

I also get $\sin(x)=\frac {\sqrt{5}-1}{2}$ as solution for the quadratic equation.You can indeed reject the other solution to the quadratic equation, $\frac {-\sqrt{5}-1}{2}$.

I don't understand your remark about cos(x).

Anyway, $\sin(x)=\frac {\sqrt{5}-1}{2}$ should give you the real solutions (I think the solutions will have to include the $\arcsin$ function, but I'm not sure, maybe someone can find a nice solution without $\arcsin$).

 

[diredragon]

I also get $\sin(x)=\frac {\sqrt{5}-1}{2}$ as solution for the quadratic equation.You can indeed reject the other solution to the quadratic equation, $\frac {-\sqrt{5}-1}{2}$.

I don't understand your remark about cos(x).

Anyway, $\sin(x)=\frac {\sqrt{5}-1}{2}$ should give you the reals solutions (I think the solutions will have to include the $\arcsin$ function, but I'm not sure, maybe someone can find a nice solution without $\arcsin$).

I made a mistake concerning cosin i figured cosx = (1 - sinx)^1/2 but that doesn't help me. The solutions of this problem somehow involves specific angles.
I have the solution but not the process of solving. It reads (π/6, π/4].
Any chance someone helping us with this?

 

[Samy_A]

I made a mistake concerning cosin i figured cosx = (1 - sinx)^1/2 but that doesn't help me. The solutions of this problem somehow involves specific angles.
I have the solution but not the process of solving. It reads (π/6, π/4].
Any chance someone helping us with this?

How can π/4 be a solution?
cos(π/4)=sin(π/4)=$\sqrt2/2$. Let's set $y=\sqrt2/2$, then your original equation becomes $\log_yy=4*log_yy$, or $1=4$.

π/6 doesn't look like a solution either (did a quick check in Excel).

 

[diredragon]

I don't understand either. It says in the problem to find all real solutions to that equation and it gives its solution as
All real solutions are in the range from pi/6 to pi/4 discluding the pi/6 and including the pi/4. I have no idea how they got to that.

 

[Samy_A]

Ow, you meant that the solution lies in that interval. Yes, that is the case, up to the 2πn terms.

 

[diredragon]

How did it get to that? How to know which interval includes the real solutions to this?

 

[Samy_A]

How did it get to that? How to know which interval includes the real solutions to this?

We know that $\sin(x)=\frac {\sqrt{5}-1}{2}=0.618 ...$
$\arcsin(0.618)=0.666 ...$ in radians, and that is approximately 38°. (I take the solution in the first quadrant, as sin(x) and cos(x) must both be positive.)
So the solution indeed lies between π/6 (30°) and π/4 (45°).

Of course, once you have a solution x, x+2πn (where n is an integer) will also be a solution.