a)
b)
a)$\displaystyle \cos x + 1 = \sin x$
Square both sides:
$\displaystyle (\cos x + 1)^2 = \sin^2 x$
$\displaystyle \cos^2 x + 2\cos x + 1 = \sin^2 x$
Use the Pythagorean identity on the RHS:
$\displaystyle \cos^2 x + 2\cos x + 1 = 1 - \cos^2 x$
Move everything to the LHS and you'll have a quadratic on cos x. Solve for cos x, and then find x. Can you take it from here?
b)$\displaystyle 2\cos 3x - 1 = 0$
Solve for cos 3x:
$\displaystyle 2\cos 3x = 1$
$\displaystyle \cos 3x = \frac{1}{2}$
Now ask yourself, the cosine of what angle(s) is 1/2? (You should know; they are all special angles!) The answers you come up with equal to 3x, which means that you'll have to divide each of the answers by 3 to solve for x.
You would have to prove this mathematically, but clearly from the graph, the first equation has solutions $\displaystyle \pi \left( {\cfrac{3}{4}\ \pm\ \cfrac{1}{4} \right)\ +\ 2\pi n\ \ \left( n \in \mathbb Z \right) $
For the second, you need $\displaystyle cos(3x)=\cfrac{1}{2}$. You should be able to do that.
Eumyang is right, although this method does lead to an extraneous answer, so you have to be careful: we have:
$\displaystyle 2cos^2 x + 2 cos x = 0$
$\displaystyle cos x \left( cos x + 1 \right) = 0$
$\displaystyle cos x = 0\ or\ -1$
However, this would lead to one set of solutions $\displaystyle \cfrac{3\pi}{2} + 2\pi n$ which actually doesn't fit the original equation. The problem arises from the fact that, given the equations
$\displaystyle (1)\ \ LHS \ = \ RHS$
$\displaystyle (2)\ \ (LHS)^2 \ = \ (RHS)^2$
we have (1) implies (2), but not (2) implies (1).
(a)
Another way...
Let (x,y) be a point on the circumference of the unit circle.
$\displaystyle cos\theta+1=sin\theta$
$\displaystyle sin\theta-cos\theta=1$
$\displaystyle \displaystyle\huge\frac{y}{r}-\frac{x}{r}=1$
$\displaystyle r=1\ \Rightarrow\ y-x=1$
By Pythagoras' theorem
$\displaystyle \sqrt{x^2+y^2}=1$
$\displaystyle \sqrt{(y-x)^2}=\sqrt{y^2+x^2}$
$\displaystyle \sqrt{y^2-2xy+x^2}=\sqrt{y^2+x^2}$
$\displaystyle xy=0\ \Rightarrow\ x=0\ or\ y=0$
Test the 4 possible angles
$\displaystyle cos(0)+1=2\ \ne\ sin(0)$
$\displaystyle cos\left(90^o\right)+1=1=sin\left(90^o\right)$
$\displaystyle cos\left(180^o\right)+1=0=sin\left(180^o\right)$
$\displaystyle cos\left(270^o\right)+1=1\ \ne\ sin\left(270^o\right)$