# Thread: Solving for unknown "x" involving trig function

1. ## Solving for unknown "x" involving trig function

I have been trying to solve 3 questions, I simply cant. Please find the attachment.

Question
13e: I am not sure whether I did it right or not
13g: After putting it in quadratic formula, i dont know how to deal with the square-root part
13h: I just cant seem to simplify the left side and solve for X

Thanks, all help is appreciated

2. Originally Posted by dennosan
13g: After putting it in quadratic formula, i dont know how to deal with the square-root part

$\displaystyle\sin x = \frac{-(-3)\pm\sqrt{(-3)^2- 4(1)(-2)}}{2(1)}$

$\displaystyle\sin x = \frac{3\pm\sqrt{9+8}}{2}$

$\displaystyle\sin x = \frac{3\pm\sqrt{17}}{2}$

$\displaystyle\sin x = \frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}$

$\displaystyle x = \sin^{-1}\left( \frac{3-\sqrt{17}}{2}\right) = \dots$

3. For problems like 13e, I always write everything in terms of sin's and cos's:

$\cot(x)\cos^{2}(x)=2\cot(x)$ implies

$\displaystyle{\frac{\cos(x)}{\sin(x)}\,\cos^{2}(x) =2\frac{\cos(x)}{\sin(x)}.}$

So we see immediately that locations where $\sin(x)=0$ are not allowed, which means $x\not=n\pi,$ for all $n\in\mathbb{Z}.$

Also, we wish to record the fact that places where $\cos(x)=0$ are a solution. So that occurs at odd integer multiples of $\pi/2,$ so at $x=(2n+1)\pi/2.$ Now that we have recorded those two facts, we may cancel all kinds of stuff. What do you get when you do that?

For 13h, I would notice that secant squared's look like tangent squared's through the Pythagorean Theorem:

$\sin^{2}(x)+\cos^{2}(x)=1$ implies, through dividing through by $\cos^{2}(x),$ that

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

So what can you do then?

4. d is wrong
wrong factorisation.