Solve Trig Equation: tanx-1 = cos2x

[TheRedDevil18]

Homework Statement

tanx-1 = cos2x

The Attempt at a Solution

I know tanx = sinx/cosx but I don't know which identity to pick for cos2x

 

[Mark44]

Homework Statement

tanx-1 = cos2x

The Attempt at a Solution

I know tanx = sinx/cosx but I don't know which identity to pick for cos2x

Pick one and see if it works.

 

[BrettJimison]

Good day TheRedDevil18

If you replace cos(2x) with 2cos²(x) - 1 ,
that will get rid of the (-1) on the left side. Then go from there...

 

[TheRedDevil18]

Good day TheRedDevil18

If you replace cos(2x) with 2cos²(x) - 1 ,
that will get rid of the (-1) on the left side. Then go from there...

sinx/cosx = 2cos^2(x)
sinx/cosx = 2(1-sin^2(x))
sinx/2cosx + sin^2(x) = 1
sinx/2cosx * 1/sin^2(x) = 1
2SinxCosx = 1
sin2x = 1
x = 45

All good?

 

[Mark44]

sinx/cosx = 2cos^2(x)
sinx/cosx = 2(1-sin^2(x))
sinx/2cosx + sin^2(x) = 1
sinx/2cosx * 1/sin^2(x) = 1

?
What are you doing here (above)?

2SinxCosx = 1
sin2x = 1
x = 45

All good?

No.
If you check your work, you'll see that 45° is not a solution.

In addition, there probably an infinite number of solutions, unless the problem specifies that you should find only the solutions in a certain interval. Since that's not stated in your post, you need to list all the solutions.

 

[Saitama]

If you check your work, you'll see that 45° is not a solution.

45° satisfies the given equation but RedDevil reached it with a wrong method as you pointed out.

 

[Mark44]

45° satisfies the given equation but RedDevil reached it with a wrong method as you pointed out.

Thanks! He didn't show the original equation in his later work, and I misremembered cos2x as cos²x in my check.

 

[verty]

I can't see how to solve this in an easy way. I get to

$sin(x) = 2 \; cos^3(x)$

and am stuck. I suspect something is wrong with this question.

 

[BrettJimison]

Remember to specify interval or say "45°+ 360°(n)" as Mark44 pointed out

 

[haruspex]

I can't see how to solve this in an easy way. I get to

$sin(x) = 2 \; cos^3(x)$

and am stuck. I suspect something is wrong with this question.

Go back to $tan(x) = 2 \; cos^2(x)$ and write $cos^2$ as $1/sec^2 = 1/(1+tan^2)$ to get a cubic in tan(x). One solution is obvious, the other two are complex.

RD18 turned sinx/2cosx + sin²(x) = 1 into sinx/2cosx * 1/sin²(x) = 1 instead of sinx/2cosx * 1/sin²x +1 = 1/sin²x, but got the right answer through sheer luck.

 

[ehild]

I can't see how to solve this in an easy way. I get to

$sin(x) = 2 \; cos^3(x)$

and am stuck. I suspect something is wrong with this question.

cos²x=1/(1+tan²x)

So you can write the equation as

tan³x+tanx-2=0,

rearrange: (tan³x-1)+(tanx-1)=0

factorize: The common factor is tanx-1.

ehild