Complex Trig Equations

[chelser13]

I am a little confused on the complex trig problems and am at a dead end on some of them. Please explain the steps on how I get to the answer!

1. (cos x/1+sin x) + (1+sin x/ cos x)

2. 6 sin^2 x - 7 six x + 2= 0

3. 6 sin ^2 x= 7- 5 cos x

Thank you!

 

[Loren]

1. (cos x/1+sin x) + (1+sin x/ cos x) <<< I think you mean cos x/(1+sin x) + (1+sin x)/cos x. Do it the same way you would do a/b + b/a...\(\displaystyle \frac{a}{b}+\frac{b}{a}=\frac{a^2}{ab}+\frac{b^2}{ab}=\frac{a^2+b^2}{ab}\)

2. 6 sin^2 x - 7 six x + 2= 0 <<< I guess you are solving for x. If it read 6a^2 - 7a + 2 = 0, could you solve for a? If so, solve your original equation for "sin x". Just like solving for a, you will get two possibilities, i.e., sin x = something or sin x = something else. Then solve for x for each of those. Gotta go to supper.

 

[soroban]

Hello, chelser13!

Here's the third one . . .
I will assume the answer are between \(\displaystyle 0\text{ and }2\pi.\)

\(\displaystyle 3)\;\;6\sin^2\!x \;=\; 7- 5\cos x\)

Replace \(\displaystyle \sin^2\!x\) with \(\displaystyle 1 - \cos^2\!x\)

And we have: .\(\displaystyle 6(1-\cos^2\!x) \;=\;7 - \cos x\)

. . which simplifies to: .\(\displaystyle 6\cos^2\!x - 5\cos x + 1 \:=\:0\)

. . which factors: .\(\displaystyle (2\cos x - 1)(3\cos x - 1) \:=\:0\)


And we have two equations to solve:

\(\displaystyle 2\cos x -1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2}\quad\Rightarrow\quad x \:=\:\tfrac{\pi}{3},\:\tfrac{5\pi}{3}\)

\(\displaystyle 3\cos x - 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{3} \quad\Rightarrow\quad x \:=\:\arccos(\tfrac{1}{3}) \:\approx\: 1.23,\;5.05\)

 

[Deleted member 4993]

I am a little confused on the complex trig problems and am at a dead end on some of them. Please explain the steps on how I get to the answer!

1. (cos x/1+sin x) + (1+sin x/ cos x)

as written - this means:

\(\displaystyle \frac{\cos(x)}{1} + \sin(x) +1 + \frac{\sin(x)}{\cos(x)}\)

I think you meant:

\(\displaystyle \frac{\cos(x)}{1+\sin(x)} = \frac{1+\sin(x)}{\cos(x)}\)

\(\displaystyle \cos^2(x) = [1+\sin(x)]^2\)

Now contonue.....
2. 6 sin^2 x - 7 six x + 2= 0

3. 6 sin ^2 x= 7- 5 cos x

Thank you!