MHB How to Solve the Trigonometric Equation 6*Sin^2(x) - 3*Sin^2(2x) + Cos^2(x) = 0?

[Igor1]

Good day :)!

Please advise how to start with the following trigonometric equation:

6*Sin^2(x) - 3*Sin^2(2x) + Cos^2(x) = 0

To be honest, I do not know what is the first steps to start with.

I have tried to start with:

5*Sin^2(x) + Sin^2(x) + Cos^2(x) - 3*Sin^2(2x) = 0

1 + 5*Sin^2(x) - 3*Sin^2(2x) = 0

but from this point I do not see the next steps to be taken.

Thank you for your help!:)

 

[MarkFL]

Good day :)!

Please advise how to start with the following trigonometric equation:

6*Sin^2(x) - 3*Sin^2(2x) + Cos^2(x) = 0

To be honest, I do not know what is the first steps to start with.

I have tried to start with:

5*Sin^2(x) + Sin^2(x) + Cos^2(x) - 3*Sin^2(2x) = 0

1 + 5*Sin^2(x) - 3*Sin^2(2x) = 0

but from this point I do not see the next steps to be taken.

Thank you for your help!:)

Hello and welcome to MHB, Igor! (Wave)

Before we proceed, are you sure the equation isn't the following:

$$6\sin^2(x)-3\sin(x)+\cos^2(x)=0$$ ?

The reason I ask, is that it seems odd that the equation would be given with like terms not combined already. :D

edit: Oops...now I see I was mistaken...so let's proceed:

$$6\sin^2(x)-3\sin^2(2x)+\cos^2(x)=0$$

The first thing I would do is use the double-angle identity to write:

$$6\sin^2(x)-3\left(2\sin(x)\cos(x)\right)^2+\cos^2(x)=0$$

$$6\sin^2(x)-12\sin^2(x)\cos^2(x)+\cos^2(x)=0$$

Now, at this point, we have a choice whether we want to convert the sines to cosines or the cosines to sines using a Pythagorean identity...which way would you choose?

 

[Igor1]

Now, at this point, we have a choice whether we want to convert the sines to cosines or the cosines to sines using a Pythagorean identity...which way would you choose?

Please let me some time to think about it. I first glance I would choose to convert the cosines to sines...

 

[MarkFL]

Please let me some time to think about it. I first glance I would choose to convert the cosines to sines...

Either choice is fine, and I would also choose to convert cosines to sines...so let's do that:

$$6\sin^2(x)-12\sin^2(x)\left(1-\sin^2(x)\right)+\left(1-\sin^2(x)\right)=0$$

Distribute:

$$6\sin^2(x)-12\sin^2(x)+12\sin^4(x)+1-\sin^2(x)=0$$

Combine like terms:

$$12\sin^4(x)-7\sin^2(x)+1=0$$

Okay, now we have a quadratic in $\sin^2(x)$...can we factor?

 

[Igor1]

Either choice is fine, and I would also choose to convert cosines to sines...so let's do that:

$$6\sin^2(x)-12\sin^2(x)\left(1-\sin^2(x)\right)+\left(1-\sin^2(x)\right)=0$$

Distribute:

$$6\sin^2(x)-12\sin^2(x)+12\sin^4(x)+1-\sin^2(x)=0$$

Combine like terms:

$$12\sin^4(x)-7\sin^2(x)+1=0$$

Okay, now we have a quadratic in $\sin^2(x)$...can we factor?

Thank you a lot I am to slow to respond...

I got the solution:

x1 = 0.6154 +/- 2Pi and 0.5335 +/- 2Pi

the factors for the quadratic equation 1/3 and 1/4.

Thank you very much for your help!:)

 

[Igor1]

Thank you a lot I am to slow to respond...

I got the solution:

x1 = 0.6154 +/- 2Pi and 0.5335 +/- 2Pi

the factors for the quadratic equation 1/3 and 1/4.

Thank you very much for your help!:)

Here is what I did further:

$$\sin^2(x)=y$$

Substitution leads to:

$$12y^2-7y+1=0$$

If we factorize this equation we get:

$$12y^2-7y+1=(3y-1)(4y-1)$$

$$y=1/3$$ and $$y=1/4$$

Further:

$$\sin^2(x)=1/3$$ and $$\sin^2(x)=1/4$$

$$x1=0.6154 \pm 2\pi$$ and $$x1=0.5235 \pm 2\pi$$

Thank you a lot for your guidance and help. It is very nice first experience here.:)

 

[MarkFL]

Thank you a lot I am to slow to respond...

I got the solution:

x1 = 0.6154 +/- 2Pi and 0.5335 +/- 2Pi

the factors for the quadratic equation 1/3 and 1/4.

Thank you very much for your help!:)

Yes, factoring works here, and we can write:

$$\left(4\sin^2(x)-1\right)\left(3\sin^2(x)-1\right)=0$$

Let's next equate each factor to zero to solve for $x$:

1.) $$4\sin^2(x)-1=0$$

$$\sin^2(x)=\frac{1}{4}$$

$$\sin(x)=\pm\frac{1}{2}$$

From this we determine:

$$x=k\pi\pm\frac{\pi}{6}=\frac{\pi}{6}\left(6k\pm1\right)$$ where $$k\in\mathbb{Z}$$

2.) $$3\sin^2(x)-1=0$$

$$\sin^2(x)=\frac{1}{3}$$

$$\sin(x)=\pm\frac{1}{\sqrt{3}}$$

From this we determine:

$$x=k\pi\pm\arcsin\left(\frac{1}{\sqrt{3}}\right)$$ where $$k\in\mathbb{Z}$$

You should find 8 roots in the interval $0\le x<2\pi$...consider the following graph:

[DESMOS=-1.094436575481077,1.301730742383198,-1.0392153769483545,1.0444083777162325]x^2+y^2=1;y=\frac{1}{2};y=-\frac{1}{2};y=\frac{1}{\sqrt{3}};y=-\frac{1}{\sqrt{3}}[/DESMOS]

You see, there are 8 points where the 4 lines intersect with the circle, corresponding to 8 angles in the interval $0\le x<2\pi$ satisfying the given equation. :D