Double-angle formulas

[Leah5467]

Hello,this is my question:

i got sinx=0,and cosx=-3/2. Cos x does not equal to the answer. =Tthe answer said x should be 0,180 and 360 degrees. Thank you!

 

[MarkFL]

Using the double-angle identity for sine, we may write:

[MATH]2\sin(x)\cos(x)+3\sin(x)=0[/MATH]
Factor:

[MATH]\sin(x)(2\cos(x)+3)=0[/MATH]
As you observed, there is no real solution for \(\cos(x)=-\dfrac{3}{2}\), and so we are left with:

[MATH]\sin(x)=0[/MATH]
And this implies (in degrees)

[MATH]x=k\cdot180^{\circ}[/MATH] where \(k\in\mathbb{Z}\)

If we are told \(0^{\circ}\le x\le360^{\circ}\)

then:

[MATH]0^{\circ}\le k\cdot180^{\circ}\le360^{\circ}[/MATH]
[MATH]0\le k\le2[/MATH]
Hence:

[MATH]x\in\left\{0^{\circ},180^{\circ},360^{\circ}\right\}[/MATH]

 

[Leah5467]

Okay thank you! But i have one question:Why do some degrees have no real solutions? Is it something that a non-mathematician could understand? If not,then i think i will just leave it. Thank you!

 

[MarkFL]

In this problem, we must consider:

[MATH]-1\le\cos(x)\le1[/MATH]
Any value outside of that range will lead to a complex solution, that is one that is not real. For problems like this, we are interested only in the real solutions.

 

[Leah5467]

i see! thank you!