I'm having trouble with these problems
Question is: "Solve the following equations for 0<X<360
1) Sin(2X)=Cos(X)
2)1+Sin(x)=Cos^2(x)
If anyone could help, I'd be really grateful.
[2] $\displaystyle 1+ \sin x= \cos^2x$
Identity: $\displaystyle cos^2 x = 1- \sin^2 x$
$\displaystyle 1+\sin x=1-\sin^2 x$
$\displaystyle \sin^2 x+ \sin x=0$
$\displaystyle \sin x(\sin x + 1)=0$
$\displaystyle \sin x=0 \ \ or \ \ \sin x = -1$
$\displaystyle x=\{0^{\circ}\:,\: 180^{\circ}\:,\:270^{\circ}\}$
It is better to factorise these and when you divide through by $\displaystyle \cos(x)$ you potentially divide by 0.
$\displaystyle 2\sin(x) \cos(x)-\cos(x)=0$
$\displaystyle \cos(x)[2\sin(x)-1]= 0$
Therefore in addition \cos(x)=0 also applies. This means that $\displaystyle x = \{90^{\circ}\:,\: 270^{\circ}\}$