Trigonometric Equations

[Marqui]

Can someone help me out with these trig problems? My PreCalculus book doesn't really explain this topic very well and I'm confused on how to solve these problems. Any sort of help would be appreciated, thanks.

Trigonometry: Find all solutions with 0<=x<2pi

1) cos(2x) + sqrt(2) = 0

2) 2sin^2 x - cosx - 1 = 0

3) 4cos^2 (2x) = 1

4) sin(2x) = -sinx

 

[wjm11]

Trigonometry: Find all solutions with 0<=x<2pi

1) cos(2x) + sqrt(2) = 0

2) 2sin^2 x - cosx - 1 = 0

3) 4cos^2 (2x) = 1

4) sin(2x) = -sinx

Hello, Marqui,

One approach, to see what’s going on, is to graph these problems.

For example, if you graph y = cos(2x) + sqrt(2), you’ll likely notice that the function never equals zero. This is because cos(2x) can only vary between +1 and –1, so if you add it to sqrt(2), which is about 1.414, the lowest value you can have is .414. So there is no solution to problem 1.

Graphing the problems will help you see solutions, but these still need to be solved using algebra and trigonometric identities. Consider the Pythagorean Trig identity sin^2(x) + cos^2(x) = 1. Rearranging this, we get sin^2(x) = 1 – cos^2(x). This might be a useful substitution in problem 2 so you can get everything into terms of cos(x) and then treat it as a quadratic equation.

I hope some of this is helpful. Please reply, showing any work you have done.

 

[soroban]

Hello, Marqui!

We're expected to know the trig values for \(\displaystyle 30^o, 45^o, 60^o\), etc.

Find all solutions with \(\displaystyle 0 \leq x \leq 2\pi\)

\(\displaystyle 1)\;\;\cos(2x) + \sqrt{2} \:=\: 0\)

We have: .\(\displaystyle \cos(2x) \:=\:-\sqrt{2}\)

Since \(\displaystyle -1 \leq \cos\theta \leq +1\), the equation has no solution.

\(\displaystyle 2)\;\; 2\sin^2\!x - \cos x - 1 \:=\: 0\)

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

We have: .\(\displaystyle 2(1-\cos^2\!x) - \cos x - 1 \:=\:0 \quad\Rightarrow\quad 2\cos^2\!x + \cos x - 1 \:=\:0\)

Factor: .\(\displaystyle (\cos x - 1)(2\cos x + 1) \:=\:0\)


\(\displaystyle \text{Therefore: }\;\begin{array}{ccccccccc}\cos x - 1 \:=\:0 & \Rightarrow & \cos x \:=\:1 & \Rightarrow & \boxed{x \:=\:0,2\pi} \\ \\[-3mm] 2\cos x + 1 \:=\:0 & \Rightarrow & \cos x \:=\:\text{-}\frac{1}{2} & \Rightarrow &\boxed{ x \:=\:\frac{2\pi}{3},\:\frac{4\pi}{3}} \end{array}\)

\(\displaystyle 3)\;\;4\cos^2\!2x \:=\: 1\)

\(\displaystyle \text{We have: }\;\cos^2\!2x \:=\:\frac{1}{4} \quad\Rightarrow\quad \cos2x \:=\:\pm\frac{1}{2}\)

\(\displaystyle \text{Then: }\;2x \;=\;\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}, \frac{11\pi}{3}\)

\(\displaystyle \text{Therefore: }\;\boxed{x \;=\;\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{4\pi}{3}, \frac{5\pi}{3},\frac{11\pi}{6}}\)

\(\displaystyle 4)\;\;\sin(2x) = -\sin x\)

\(\displaystyle \text{We have: }\;\sin(2x) + \sin x \;=\;0 \quad\Rightarrow\quad 2\sin x\cos x + \sin x \:=\:0\)

\(\displaystyle \text{Factor: }\;\sin x(2\cos x + 1) \:=\:0\)


\(\displaystyle \text{Therefore: }\;\begin{array}{ccccccc}\sin x \:=\:0 & \Rightarrow &\boxed{x \:=\:0,\pi,2\pi} \\ \\[-3mm]2\cos x + 1 \:=\:0 & \Rightarrow & \cos x \:=\:\text{-}\frac{1}{2} & \Rightarrow & \boxed{x \:=\:\frac{2\pi}{3},\frac{4\pi}{3}} \end{array}\)

 

[Marqui]

Thanks a lot guys. You broke it down and everything, I understand the problems now! Thanks again.