a) $\displaystyle Sec^2(x) - 2Tan(x) = 4$
b) $\displaystyle Cos(x) + 1 = Sin(x)$
c) $\displaystyle 2Cos(3x) - 1 = 0$
I'll do the first one.
$\displaystyle \sec^2\,x - 2\tan\,x = 4$
Use the Pythagorean identity $\displaystyle 1 + \tan^2\,\theta = \sec^2\,\theta$:
$\displaystyle \begin{aligned}
1 + \tan^2\,x - 2\tan\,x &= 4 \\
\tan^2\,x - 2\tan\,x - 3 &= 0
\end{aligned}$
Hey, this looks like a quadratic! Not only that, it's factorable! Factor this and set each factor equal to zero:
$\displaystyle \begin{aligned}
\tan^2\,x - 2\tan\,x - 3 &= 0 \\
(\tan\,x - 3)(\tan\,x + 1) &= 0 \\
\end{aligned}$
...
I'll do the second one.
$\displaystyle \cos x+1=\sin x$
Note that the identity
$\displaystyle (1-\cos x)(1+\cos x)=1-\cos^2x=sin^2x$
stands for all possible real numbers $\displaystyle x$.
And by squaring up both sides of equation (2), we have
$\displaystyle \sin^2 x=(1+\cos x)(1+\cos x)$
Therefore it's possible to cancel the term \sin^2x:
$\displaystyle (1-\cos x)(1+\cos x)=(1+\cos x)(1+\cos x)\Rightarrow 2\cos x(1+\cos x)=0$
Therefore
$\displaystyle \cos x=0, \sin x=1\;\text{or}\;\cos x=-1, \sin x=0$
That being said,
$\displaystyle x=90^\circ+360^\circ\cdot k\;\text{where $k$ is an integer} $
or
$\displaystyle x=180^\circ+360^\circ\cdot k\;\text{where $k$ is an integer} $
You're too late. The OP reposted b) and c) in this thread: http://www.mathhelpforum.com/math-he...ns-153161.html . No fault on your part -- she shouldn't have done that.