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

In summary: Yes, factoring works here, and we can write:\left(4\sin^2(x)-1\right)\left(3\sin^2(x)-1\right)=0Let'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
  • #1
Igor1
4
0
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!:)
 
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  • #2
Igor said:
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?
 
  • #3
MarkFL said:
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...
 
  • #4
Igor said:
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?
 
  • #5
MarkFL said:
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!:)
 
  • #6
Igor said:
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.:)
 
  • #7
Igor said:
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
 

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

What is a trigonometric equation?

A trigonometric equation is an equation that contains one or more trigonometric functions (such as sine, cosine, or tangent) and one or more variables. The goal of solving a trigonometric equation is to find the values of the variables that make the equation true.

What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, and tangent. These functions represent the ratio of two sides of a right triangle, with respect to a given angle. Other common trigonometric functions include cotangent, secant, and cosecant, which are the reciprocals of sine, cosine, and tangent respectively.

How do you solve a trigonometric equation?

To solve a trigonometric equation, you must use algebraic techniques to isolate the variable and then use trigonometric identities and properties to simplify the equation. You may also need to use the unit circle and the special angles (30°, 45°, and 60°) to find the exact solutions.

What are the general solutions of a trigonometric equation?

The general solutions of a trigonometric equation are all possible solutions that make the equation true. These solutions include the initial solutions found using algebraic techniques, as well as any additional solutions that can be found by adding multiples of the period of the function.

What is the difference between an identity and an equation?

An identity is a statement that is true for all values of the variables, while an equation is a statement that is only true for specific values of the variables. In other words, an identity is always true, while an equation may or may not have solutions that make it true.

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