# Thread: Solving Trig Equations Help

1. ## Solving Trig Equations Help

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. Originally Posted by LexiLou
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.
Hi LexiLou,

$\sin 2x=2\sin x \cos x$, so

[1] $\sin 2x=\cos x$

$2\sin x \cos x=\cos x$

$2\sin x = 1$

$\sin x =\frac{1}{2}$

$x=\{30^{\circ}\:,\:150^{\circ}\}$

3. [2] $1+ \sin x= \cos^2x$

Identity: $cos^2 x = 1- \sin^2 x$

$1+\sin x=1-\sin^2 x$

$\sin^2 x+ \sin x=0$

$\sin x(\sin x + 1)=0$

$\sin x=0 \ \ or \ \ \sin x = -1$

$x=\{0^{\circ}\:,\: 180^{\circ}\:,\:270^{\circ}\}$

4. Originally Posted by masters
Hi LexiLou,

$\sin 2x=2\sin x \cos x$, so

[1] $\sin 2x=\cos x$

$2\sin x \cos x=\cos x$

$2\sin x = 1$

$\sin x =\frac{1}{2}$

$x=\{30^{\circ}\:,\:150^{\circ}\}$
It is better to factorise these and when you divide through by $\cos(x)$ you potentially divide by 0.

$2\sin(x) \cos(x)-\cos(x)=0$

$\cos(x)[2\sin(x)-1]= 0$

Therefore in addition \cos(x)=0 also applies. This means that $x = \{90^{\circ}\:,\: 270^{\circ}\}$

5. Originally Posted by e^(i*pi)
It is better to factorise these and when you divide through by $\cos(x)$ you potentially divide by 0.

$2\sin(x) \cos(x)-\cos(x)=0$

$\cos(x)[2\sin(x)-1]= 0$

Therefore in addition \cos(x)=0 also applies. This means that $x = \{90^{\circ}\:,\: 270^{\circ}\}$
Thanks, e^(i*pi). Got in a hurry! Thanks for cleaning up after me.

6. Thanks so much, that helped a lot.