# Thread: Having a hard time working out trig equations

1. ## Having a hard time working out trig equations

My instructor emailed us a study guide with an answer key, but some of the problems aren't worked out. There are only examples of the easy problems in the book, but not the hard ones (just like every fricken problem).

a: 2sinx = the square root of 3

Answer: π/3 + 2kπ; 2π/3 + 2kπ
I'm pretty sure I have to find the arcsin of the square root of 3, but I'm not sure when I'm supposed to multiply that by 2.

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

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

Answer: 0.878 + kπ; 2.264 + kπ

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

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

Answer: (the square root of 3)/(2)
For this problem, I'm just having a hard time converting the degrees to xy coordinates. I know how to solve problems like cos(30°)cos(45°) + cos(30°)sin(45°) because I have the xy coordinates memorized for it. I'm not sure how to actually use a formula and find them however.

Let me know if there's something I need to thoroughly explain.

My instructor emailed us a study guide with an answer key, but some of the problems aren't worked out. There are only examples of the easy problems in the book, but not the hard ones (just like every fricken problem).

a: 2sinx = the square root of 3

Answer: π/3 + 2kπ; 2π/3 + 2kπ
I'm pretty sure I have to find the arcsin of the square root of 3, but I'm not sure when I'm supposed to multiply that by 2.

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

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

Answer: 0.878 + kπ; 2.264 + kπ

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

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

Answer: (the square root of 3)/(2)
For this problem, I'm just having a hard time converting the degrees to xy coordinates. I know how to solve problems like cos(30°)cos(45°) + cos(30°)sin(45°) because I have the xy coordinates memorized for it. I'm not sure how to actually use a formula and find them however.

Let me know if there's something I need to thoroughly explain.

Yes, to do the question c) you need the calculator.

To the question e), see: Product and Sum Formulas

Welcome to Math Help Forum!
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a: 2sinx = the square root of 3

Answer: π/3 + 2kπ; 2π/3 + 2kπ
$2\sin x = \sqrt3$

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

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

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

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

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

$\Rightarrow x = -\frac{\pi}{12}-\frac53+\frac{k\pi}{3}$ (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:
$1-\frac{1}{\cos2x}=6.438$

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

$\Rightarrow \cos2x = -0.1839$
Can you complete it now?

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

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

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

$\cos x = 4$ (impossible) or $\cos x = -1$
You should be able to complete this now

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

Answer: (the square root of 3)/(2)
$\cos20^o = \sin 70^o$ and $\sin(-10^o) = -\sin10^o$. So we can say:
$\cos20^o\cos10^o + \cos70^o\sin(-10^o)$
$=\sin70^o\cos10^o-\cos70^o\sin10^o$

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

$=\sin60^o$

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