Recent content by Igor1

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!:)

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

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) -...