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...
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!:)
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) -...