# Math Help - solving trig equations

1. ## solving trig equations

I have a few problems that I just can't figure out how to deal with at all.

1. 1-sin(t) = sqrt3(cos t)

2. cos(x) - sin(x) = 1

(I know that one's really simple, but I don't know how to set it up.)

3. 2tan(t) - sec^2(t) = 0

4. tan(x) + sec(x) = 1

5. cot(a) + tan(a) = csc(a)sec(a)

6. sec^5(x) = 4sec(x)

The section we're on is factoring trig equations, but my basic factoring skills have gone all dry and I can't remember how to do it simply. If I knew how to factor these I could get the answers, so the factoring is mainly what I need help on--or if it's not factoring, then how to set it up so I can find all the angles within the equation.

2. 1. $1-\sin(t) = \sqrt{3}\cos(t)$

$(1-\sin(t))^2 = ( \sqrt{3}\cos(t))^2$

$\sin^2(t) +2 \sin(t) +1 = 3\cos^2(t)$

$\sin^2(t) +2 \sin(t) +1= 3(1-\sin^2(t))$

$\sin^2(t) +2 \sin(t) +1+3\sin^2(t) -3= 0$

$4\sin^2(t) +2 \sin(t) -2= 0$

$2\sin^2(t) + \sin(t) -1= 0$

$( \sin(t) -1)(2 \sin(t) +1) = 0$

you could take it from here...

2. $\cos(x) - \sin(x) = 1$

$( \cos(x) - \sin(x))^2 = 1^2$

$\cos^2(x) - 2\cos(x)\sin(x) + \sin^2(x) = 1$

$1- \sin^2(x) - 2\cos(x)\sin(x) + \sin^2(x) = 1$

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

you could take it from here...

3. $2\tan(t) - \sec^2(t) = 0$

$2\tan(t) - (1+\tan^2(t)) = 0$

$2\tan(t) - 1-\tan^2(t)) = 0$

$\tan(t) = 1$

you could take it from here...

4. $\tan(x) + \sec(x) = 1$

$(\tan(x) + \sec(x))^2 = 1^2$

$\tan^2(x) + 2\tan(x)\sec(x) + \sec^2(x) = 1$

$\tan^2(x) + 2\tan(x)\sec(x) + 1+\tan^2(x) = 1$

$2\tan^2(x) + 2\tan(x)\sec(x) = 0$

$2 \left( \frac{\sin^2(x)}{cos^2(x)} + \frac{\sin(x)}{\cos^2(x)} \right)= 0$

$2 \left( \frac{\sin^2(x)+\sin(x)}{cos^2(x)} \right)= 0$

you could take it from here...

5. $\cot(a) + \tan(a) = \csc(a) \sec(a)$

$\frac{\cos(a)}{\sin(a)} + \frac{\sin(a)}{\cos(a)} = \frac{1}{\sin(a)} \times \frac{1}{cos(a)}$

$\frac{\cos^2(a)}{\sin(a)\cos(a)} + \frac{\sin^2(a)}{\cos(a)\sin(a)} = \frac{1}{\sin(a)\cos(a)}$

$\frac{1}{\sin(a)\cos(a)}= \frac{1}{\sin(a)\cos(a)}$

you could take it from here...

6. $\sec^5(x) = 4\sec(x)$

$\sec^4(x) = 4$

$\sec(x) = \sqrt[4]{4}$

you could take it from here...