Solving Trigonometry Equations and Problems with Examples and Tips

Solve $\sec (x) = \tan (x)$
I have been trying to do this problem: Solve $\sec (x) = \tan (x), 0 \leq x < 2 π$ I started by rewriting $\sec (x) as \frac{1}{\cos (x)}$ . I then rewrote $\tan (x) to get \frac{\sin (x)}{\cos (x)}$ Therefore: $\frac{1}{\cos x} = \frac{\sin x}{\cos x}$ Therefore: $1 = \sin x, x = \frac{π}{2}$ However, this is not a solution to the original problem because both $y = \tan (x) and y = \sec (x)$ have asymptotes as $x = \frac{π}{2}$

Kendall Holder 2022-01-23

$\sin (x) + p \cos (x) = 2 p, p \in R$ has solutions for p=?

Jamarion Kerr 2022-01-23

For proving
$\frac{16}{\cos 4 x + 7} = \frac{1}{\sin^{4} x + \cos^{2} x} + \frac{1}{\sin^{2} x + \cos^{4} x}$
I tried to use that:
$\sin^{4} x + \cos^{4} x = \sin^{4} x + 2 \sin^{2} x \cos^{2} x + \cos^{4} x - 2 \sin^{2} x \cos^{2} x$
$= {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x$
$= 1^{2} - {\frac{12}{2 \sin x \cos x}}^{2}$
$= 1 - {\frac{12}{\sin}}^{2} (2 x)$
$= 1 - \frac{12}{\frac{1 - \cos 4 x}{2}}$
$= \frac{34}{+} \frac{14}{\cos 4} x$

pozicijombx 2022-01-23

prove that $2 \arctan (\csc \arctan x - \tan a r o t x) = \arctan x$

Jenny Branch 2022-01-23

How to prove : $\cos^{3} 2 θ + 3 \cos 2 θ = 4 (\cos^{6} θ - \sin^{6} θ)$

jelentetvq 2022-01-23

How is the derivative of $x \cos x \sin x \frac{12}{\sin 2} x + x \cos 2 x$ ?
I tried using the product rule on $x \cos x$ and got:
$(x \cos x)' = - x \sin x + \cos x$
then I used the product rule to differentiate xcosxsinx and got:
$(x \cos x \sin x)' = (- x \sin x + \cos x) \sin x + (x \cos x \cos x) = - x \sin^{2} x + \sin x \cos x + x \cos^{2} x$

Jayleen Sanders 2022-01-23

Proving $\int_{0}^{\frac{π}{4}} \sin^{n} (x) d x > \frac{1}{2^{\frac{n}{2}} (n + 2)}$

Danitat6 2022-01-23

Evaluating $\sum_{k = 1}^{n - 1} k \cos (k x)$

pozicijombx 2022-01-23

Solve $y = \frac{13}{\sin x + [\sin x + [\sin x]]}$ & $[y + [y]] = 2 \cos x$
$\sin x \in (π, 2 π)$
$y = \frac{13}{\sin x + [\sin x + [\sin x]]}$
y=-1
$[- 1 + [- 1]] = 2 \cos x$
$\cos x = - 1$ which is possible at $x = π$ hence NO SOLUTION $\sin x \in \frac{π}{2}$
$y = \frac{13}{\sin x + [\sin x + [\sin x]]}$
y=1
$[1 + [1]] = 2 \cos x$
$\cos x = 1$ which is possible at x=0 hence NO SOLUTION
Similary if $\sin x \in (0, π) - \frac{π}{2}$
$y = \frac{1}{3} [\sin x + [\sin x + [\sin x]]]$
y=0
$[0 + [0]] = 2 \cos x$
$\cos x = 0$ which is possible at $x = \frac{π}{2}$ hence NO SOLUTION

veceriraby 2022-01-23

Evaluating $\int_{- \frac{π}{4}}^{\frac{π}{4}} {(\sin x + 2 \cos x)}^{3} d x$

Duncan Reed 2022-01-23

Evaluate and simplify: $lim_{x \to 0} \frac{\sin 2 x}{3 x}$
So, by direct substitution, it is indeterminate. I got the derivative of this functio,n and now Im

Danitat6 2022-01-23

General solution of $\tan (2 x) \tan (x) = 1$

2alr8w 2022-01-23

Simplifying $\sqrt{\frac{{(a^{2} \cos^{2} t + b^{2} \sin^{2} t)}^{3}}{{(b^{2} \cos^{2} t + a^{2} \sin^{2} t)}^{3}}}$

Arely Briggs 2022-01-23

Proving $\tan (x + \frac{π}{3}) = \frac{4 \tan x + \sqrt{3} \sec^{2} x}{\sec^{2} x - 4 \tan^{2} x}$

Miguel Davenport 2022-01-23

$x + y = 120^{\circ}, x \geq 0, y \geq 0$ . max and min of $\sin x + \sin y$

spiderifilms6e 2022-01-23

Prove $\sin (a) < a < \tan (a)$ when $0 < a < \frac{π}{2}$

Jacquelyn Sanders 2022-01-23

Does $A \sin b = B \sin b + C \cos b$ imply C=0?

Tori Hines 2022-01-23

How to use Liouville's theorem to prove the formula of $\frac{π}{\sin π z}$

Tori Hines 2022-01-23

Where is the angle for $- \frac{7 π}{8}$ in a unit circle?
I am having difficulty finding the angle for $- \frac{7 π}{8}$ in a unit circle. I have found the coordinates for $\frac{π}{8}$ and have to use the symmetry to locate the coordinates for $- \frac{7 π}{8}$ . Is the angle just below the (1,0) and symmetry to $\frac{π}{8}$ ?

Turnseeuw 2022-01-23

Differentiate $\sqrt{\frac{1 + \sin x}{1 - \sin x}}$

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