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:

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

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:

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

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

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

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

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:

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

Distribute:

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

Combine like terms:

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

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:

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

Distribute:

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

Combine like terms:

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

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:

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

Substitution leads to:

\(\displaystyle 12y^2-7y+1=0\)

If we factorize this equation we get:

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

\(\displaystyle y=1/3\) and \(\displaystyle y=1/4\)

Further:

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

\(\displaystyle x1=0.6154 \pm 2\pi\) and \(\displaystyle x1=0.5235 \pm 2\pi\)

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:

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

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

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

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

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

From this we determine:

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

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

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

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

From this we determine:

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

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