Solving Trigonometry Equations and Problems with Examples and Tips

Solve: $3 \sin 2 x + 4 \cos 2 x - 2 \cos x + 6 \sin x - 6 = 0$
My Try
$6 \sin x \cos x + 4 (\cos^{2} x - \sin^{2} x) - 2 \cos x + 6 \sin x - 6 = 0$

Madeline Lott 2021-12-31

I tried to solve the following problem; but I couldnt.

widdonod1t 2021-12-31

How does $\frac{4 π x + \cos (4 π x) \sin (4 π x)}{8 π} + C$ become $\frac{\sin (8 π x) + 8 π x}{16 π} + C$ ?

elvishwitchxyp 2021-12-31

Finding the derivative of $\cos (\arcsin x)$

Kaspaueru2 2021-12-31

Basically, write $\cos 4 x$ as a polynomial in $\sin x$ .
I've tried the double angles theorem and $\cos 2 x = \cos^{2} x - \sin^{2} x$ . I'm still having trouble right now though.

crealolobk 2021-12-31

Find the sationary points of the curve and their nature for the equation $y = e^{x} \cos x$ for $0 \leq x \leq \frac{π}{2}$
I derived it and got $e^{x} (- \sin x + \cos x) = 0$
$e^{x}$ has no solution but I don't know how to find the x such that $- \sin x + \cos x = 0$

Wanda Kane 2021-12-31

How to integrate hyperbolic function $\frac{1 - \sin (x)}{1 + \sin (x)}$

David Young 2021-12-31

What is general solution of $\frac{\cos 5 x \cos 3 x - \sin 3 x \sin x}{\cos 2 x} = 1$
1) $\frac{k π}{3}$
2) $\frac{k π}{2}$
3) $\frac{2 k π}{5}$
4) $\frac{2 k π}{3}$
The numerator of the fraction is $\cos (5 x + 3 x)$ . so I should find general solution of $\cos 8 x = \cos 2 x$ . I'm not sure how to do it, I can write $\cos 8 x$ in term of $\cos 2 x$ :
$\cos 8 x = 2 \cos^{2} 4 x - 1 = 2 {(2 \cos^{2} 2 x - 1)}^{2} - 1$
After substituting it in the equation and using $\cos 2 x = t$ we have degree four equation

Jessie Lee 2021-12-30

Evaluating $lim_{x \to 0} (\frac{1}{\sin (x)} - \frac{1}{x})$ without L'Hôpital's rule
The only thing I can think of using is the basic identity
$lim_{x \to 0} (\frac{\sin x}{x}) = 1$
but I can't reduce the original problem down to a point where I can apply this identity.

Marla Payton 2021-12-30

Find the general solution of equation $\cot x + \tan x = 2$
My approach:
$\cot x + \tan x = \frac{(1 + \tan^{2} x)}{\tan x} = 2$
$\sec^{2} x + \tan x = 2$
$\frac{1}{(\sin x \cos x)} = 2$
$\sin^{2} x + \cos^{2} x = 2 \sin x \cos x$
${(\sin x - \cos x)}^{2} = 0$
$\sin x = \cos x$
So, $x = n π + (\frac{π}{4})$ . This was my answer. But the answer given is $x = 2 n π \pm \frac{π}{3}$

hadejada7x 2021-12-30

Without using LHospitals rule, what is $lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}}$ ?

Tiffany Russell 2021-12-30

how to show that $\csc x - \csc (\frac{π}{3} + x) + \csc (\frac{π}{3} - x) = 3 \csc 3 x$ ?
My attempt:
$L H S = \csc x - \csc (\frac{π}{3} + x) + \csc (\frac{π}{3} - x)$
$= \frac{1}{\sin x} - \frac{1}{\sin (\frac{π}{3} + x)} + \frac{1}{\sin (\frac{π}{3} - x)}$
$= \frac{\sin x \sin (\frac{π}{3} + x) + \sin (\frac{π}{3} + x) \sin (\frac{π}{3} - x) - \sin (\frac{π}{3} - x) \sin x}{\sin x \sin (\frac{π}{3} + x) \sin (\frac{π}{3} - x)}$
$= \frac{4}{\sin 3 x} (\sin x \sin (\frac{π}{3} + x) + \sin (\frac{π}{3} + x) \sin (\frac{π}{3} - x) - \sin (\frac{π}{3}) \sin x)$
$= \frac{4}{\sin 3 x} (\sin x \sin (\frac{π}{3} + x) - \sin (\frac{π}{3} - x) \sin (\sin x - \sin (\frac{π}{3} + x)))$
$= \frac{4}{\sin 3 x} (\sin x \sin (\frac{π}{3} + x) - \sin (\frac{π}{3} - x) (2 \sin \frac{- π}{6} \cos (x + \frac{π}{6})))$

zeotropojd 2021-12-30

Why does $\sum_{i = 1}^{k} \sin (i \frac{2 π}{k}) = 0$ for integers k

hadejada7x 2021-12-30

Any trick for evaluating ${(\frac{\sqrt{3}}{2} \cos θ + \frac{i}{2} \sin (θ))}^{7}$ ?
Expressions of the form ${(a \cos (θ) + b i \sin (θ))}^{n}$ come up from time to time in applications of complex analysis, but to my knowledge the De Moivre's formula can only be applied with a=b. Is there some trick to deal with the case of $a \neq b$ , for example when the expression is ${(\frac{\sqrt{3}}{2} \cos θ + \frac{i}{2} \sin θ)}^{7}$ ?

pierdoodsu 2021-12-30

Derivative of $\cos^{- 1} \sqrt{\frac{1 + x}{2}}$ using substitution

Painevg 2021-12-28

Number of solutions of $\cos^{5} x + \cos^{5} (x + \frac{2 π}{3}) + \cos^{5} (x + \frac{4 π}{3}) = 0$

Kathleen Rausch 2021-12-27

Let the integral be:
$\int \sin^{4} (x) \cos^{3} (x) d x$
I have to integrate this function by changing the variable. I'm trying: $u = \sin (x)$ and so $d u = \cos (x) d x$ . By rewriting the integral I get:
$\int u^{4} \cos^{2} (x) d u$
But I'm stuck here because I'm not sure there should be any expression with x left in the integral.Also I know the final answer is :
$\int \sin^{4} x \cos^{3} x d x = - \frac{\sin^{7} x}{7} + \frac{\sin^{5} x}{5}$

Nicontio1 2021-12-26

The counterpart questions for sine and tangent can be handled as follows:
If $\frac{\sin A}{2} = \frac{\sin B}{3} = \frac{\sin C}{7}$ we can rule out triangle because by the Sine Rule a=2k, b=3k, c=7k $\Rightarrow a + b < c$
If $\frac{\tan A}{2} = \frac{\tan B}{3} = \frac{\tan C}{7}$ , we can see that a triangle will be made as $\tan A = 2 k, \tan B = 3 k, \tan C = 7 k$ , when inserted in the identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C \Rightarrow k = \sqrt{\frac{2}{7}}$

Carla Murphy 2021-12-26

Prove $- 4 \sin^{4} (\frac{12}{x}) = - 4 \sin^{2} (\frac{12}{x}) + \sin^{2} x$
I want to prove :
$- 4 \sin^{4} (\frac{x}{2}) = - 4 \sin^{2} (\frac{x}{2}) + \sin^{2} x$
I try to factoring :
$- 4 \sin^{2} (\frac{x}{2}) + \sin^{2} x = (\sin x + 2 \sin (\frac{x}{2})) (\sin x - 2 \sin (\frac{x}{2}))$
but I can't simplify it to left hand side.

Salvatore Boone 2021-12-21

A box with a square base and no top is to be made from a square piece of cardboard by cutting 4-in. squares from each corner and folding up the sides. The box is to hold $100 \in^{3}$ . What size cardboard piece is required?

The majority of Trigonometry Math problems are quite easy to solve when you have the list of answers that help you see the logic and reasoning. Looking at various examples will help you to see how certain questions are related to provided solutions.

Remember that Trigonometry requires patience and analysis. By following your analysis skills, you will help yourself find the answers. If something remains unclear, start with your exploration of examples again and go step-by-step.

Precalculus

Practice Exercises for Trigonometry