Hello jamesbrad288

Welcome to Math Help Forum! Originally Posted by

**jamesbrad288** ...

a: 2sinx = the square root of 3

Answer: π/3 + 2kπ; 2π/3 + 2kπ

$\displaystyle 2\sin x = \sqrt3$

$\displaystyle \Rightarrow \sin x = \frac{\sqrt3}{2}$

There are two values of $\displaystyle x$ between $\displaystyle 0$ and $\displaystyle \pi$ that satisfy this equation: $\displaystyle x=\frac{\pi}{3}$ and $\displaystyle x=\frac{2\pi}{3}$. These values then repeat every $\displaystyle 2\pi$. So the solutions are:

$\displaystyle x = \frac{\pi}{3}+2k\pi;\; \frac{2\pi}{3} + 2k\pi$

b: tan(3x + 5) = -1 (not sure if I'm supposed to stick -1 into the arctan(3(-1) + 5) or if I'm supposed to find the arctan(-1) and stick that in tan(3x + 5))

Answer: -π/13 - 5/3 + kπ/3

$\displaystyle \tan(3x+5) = -1$

$\displaystyle \Rightarrow 3x+5 = -\frac{\pi}{4}+k\pi$, since values of tan repeat every $\displaystyle \pi$.

$\displaystyle \Rightarrow x = -\frac{\pi}{12}-\frac53+\frac{k\pi}{3}$ (There's a typo in the answer you gave.)

c: 1 - sec(2x) = 6.438 (I'm pretty sure I need my calculator for this one)

Answer: 0.878 + kπ; 2.264 + kπ

It's easiest to turn it around to a cosine:$\displaystyle 1-\frac{1}{\cos2x}=6.438$

$\displaystyle \Rightarrow \frac{1}{\cos2x}=-5.438$

$\displaystyle \Rightarrow \cos2x = -0.1839$

Can you complete it now?

d: cosx^2 - 3cosx = 4 (cosx^2 is cosx square in case that was misleading)

Answer: π + 2kπ

Re-arrange as a quadratic in $\displaystyle \cos x$:$\displaystyle \cos^2x-3\cos x -4=0$

$\displaystyle (\cos x -4)(\cos x +1)=0$

$\displaystyle \cos x = 4$ (impossible) or $\displaystyle \cos x = -1$

You should be able to complete this now

e: cos(20°)cos(10°) + cos(70°)sin(-10°)

Answer: (the square root of 3)/(2)

$\displaystyle \cos20^o = \sin 70^o$ and $\displaystyle \sin(-10^o) = -\sin10^o$. So we can say:

$\displaystyle \cos20^o\cos10^o + \cos70^o\sin(-10^o)$$\displaystyle =\sin70^o\cos10^o-\cos70^o\sin10^o$

$\displaystyle =\sin(70^o-10^o)$

$\displaystyle =\sin60^o$

$\displaystyle =\frac{\sqrt3}{2}$

Grandad