# inverse trigonometric equations and inequalities

Sorry for the list, wasn't sure if i should post each question in a separate post, but they are all under the same section and chapter. I don't expect for all questions to be answered, but any help on any of them is appreciated, thanks. I made sure all arithmetic calculations were correct. I didn't finish the problems completely, but i must be on the right track first to finish them (solve for x). ill be editing the post along as i find my own answers

(for all questions, find all solutions, solve for x)

# 1.

$$2sin^2(x) + (sin(2x)) = 0$$

use identity, sin(2x) = 2sin(x)cos(x)


$$2sin^2(x) + (2sin(x)cos(x)) = 0$$

factor


$$(2sin(x))(sin(x)+cos(x)) = 0$$

$$sin(x) = 0$$

$$sin(x) = -cos(x)$$ $$\frac{sin(x)}{sin(x)} = -\frac{cos(x)}{sin(x)}$$ $$cot(x) = -1$$

sin(x)=0, cot(x)=-1 ... correct?


# 2.

$$6cos^2(\frac{x}{2})-7cos(\frac{x}{2})+2 = 0$$

cant factor, use quadratic formula ?


$$A= 6, B = -7, C = 2$$ $$\frac{7\pm \sqrt{1}}{12}$$

$$cos(\frac{x}{2}) = \frac{1}{2}$$ $$cos(\frac{x}{2}) = \frac{2}{3}$$

correct?


# 3.

$$cos^2(x)-0.2sin(x) = 0.9$$

use identity


$$(1-sin^2(x))-\frac{1}{5}sin(x) = \frac{9}{10}$$ $$-sin^2(x)-\frac{1}{5}sin(x) = \frac{9}{10} - \frac{10}{10}$$ $$-sin^2(x)-\frac{1}{5}sin(x) + \frac{1}{10} = 0$$

negate to factor


$$sin^2(x)+\frac{1}{5}sin(x) - \frac{1}{10} = 0$$

cant factor, use quadratic formula


$$A=1,B=\frac{2}{10},C=-\frac{1}{10}$$ $$sin(x) = \frac{-\frac{2}{10}\pm \sqrt{\frac{4}{100}+(\frac{4}{10})}}{2}$$

$$sin(x) = \frac{-\frac{2}{10}\pm \sqrt{\frac{11}{25}}}{2}$$

$$sin(x) = 0.231662479$$ $$sin(x) = -0.431662479$$

correct?


# 4.

$$4sin^2(2x) = sin(2x) + 3$$

use identities


$$(4)(2sin(x)cos(x))^2 = 2sin(x)cos(x) + 3$$

$$16sin^2(x)cos^2(x) - 2sin(x)cos(x) - 3 = 0$$

factor-able


$$(2sin(x)cos(x)-1)(8sin(x)cos(x)+3)$$

i wont have sin(x) or cos(x) by itself unless i divide


$$sin(x) = \frac{1}{2cos(x)}$$

$$sin(x) = \frac{-3}{8cos(x)}$$

incorrect ... ?


# 5.

$$cos(x)csc(x) = cot^2(x)$$

identities


$$\cos(x)\frac{1}{sin(x)} = cot^2(x)$$

$$cot(x) = cot^2(x)$$

$$1 = cot(x)$$

correct?


## 1 Answer

(Partial answers)

General comment: It seems that the questions ask for $x$, perhaps in some interval like $[0,2\pi)$, and perhaps without restriction. Nowhere have you mentioned what the possible values of $x$ are.

1) We have the possibility $\sin x=0$, giving $x=n\pi$, where $n$ ranges over the integers. For the possibility $\sin x+\cos x=0$, which you left undealt with, transform perhaps to $\tan x=-1$. That has solutions $x=\frac{3\pi}{4}$ and $x=-\frac{3\pi}{4}$. All solutions are given by $x=2n\pi \pm \frac{3\pi}{4}$.

2) Not that it matters much, but one can factor. You have obtained the right values for $\cos(x/2)$. That in principle tells you all possible values of $x/2$, and then of $x$. For $\cos(x/2)$, there are the solutions $x/2=2n\pi\pm \frac{\pi}{3}$. For $\cos(x/2)=\frac{2}{3}$, the expressions for $x/2$ are not nice.

4) You went to too much trouble. We have $4\sin^2(2x)-\sin(2x)-3=0$, already a quadratic. Factor as $(4\sin(2x)+3)(\sin(2x)-1)=0$.

But one can finish from where you got to, using $2\sin x\cos x=\sin(2x)$.

5) Don't forget about $\cot x=0$.

• #1) i came out with cot(x) = -1, but i can also get tan(x) = -1. are both the same answer? #5) why cot(x) = 0 ?
– hit
May 10 '14 at 3:45
• $\cot x=-1$ says the same thing as $\tan x=-1$. I used $\tan$ because it is more familiar. For Question 5, recall that the equation $w=w^2$ has two solutions, $0$ and $1$. May 10 '14 at 3:48
• #4. i still come out with the same end if i use identities, and get sin(x) on one side. do i just divide end angles by 2 using sin(2x)?
– hit
May 10 '14 at 4:14
• Dividing as you did does not get you to where you want to go, at least not quickly. If you find a general expression for the angle $2x$, then you divide the expression by $2$ to get a general expression for $x$. May 10 '14 at 4:25