Trigonmetric equations help

[kelleynicole30]

Find the solutions of the equation that are in the interval [0,2?). Please show steps (especially rearranging the equation to solve it. Thanks for helping me. I know the solutions are 11?/6, ?/2 for the first problem and 7?/6, 11?/6 for the second problem.


1A) 1-sint=(?3)cost
B) 2cos?+tan?=sec?

 

[stapel]

Please show steps (especially rearranging the equation to solve it.

We already know how to do these, so we don't really need the practice. And you already have worked examples in your textbook and in your class notes, so another worked example or two probably isn't going to make much difference in your understanding. (I'm assuming you're wanting to learn how to do this on your own, rather than hoping somebody will just "do" your homework for you.)

So please reply with your thoughts and efforts so far, even if you're fairly certain that you're on the wrong track. You might actually be on the right track, and just needing a little nudge. But we can't know that until we can "see" where you are.

Please be complete. Thank you!

Eliz.

 

[Deleted member 4993]

Find the solutions of the equation that are in the interval [0,2?). Please show steps (especially rearranging the equation to solve it. Thanks for helping me. I know the solutions are 11?/6, ?/2 for the first problem and 7?/6, 11?/6 for the second problem.


1A) 1-sint=(?3)cost

$\displaystyle \sqrt{3}\cdot cos(t) + 1\cdot sin(t)\, = \, 1$

\(\displaystyle \sqrt{\sqrt{3}^2 + 1^2}\cdot[{\frac{\sqrt{3}}{\sqrt{\sqrt{3}^2 + 1^2}}\cdot cos(t) + \frac{1}{\sqrt{\sqrt{3}^2 + 1^2}}sin(t)]\, = \, 1\)

\(\displaystyle 2\cdot[{\frac{\sqrt{3}}{2}\cdot cos(t) + \frac{1}{2}sin(t)]\, = \, 1\)

$\displaystyle [cos(\frac{\pi}{6}+2\cdot n \cdot \pi)\cdot cos(t) + sin(\frac{\pi}{6}+2\cdot n \cdot \pi)\cdot sin(t)]\, = \, \frac{1}{2}$

continue from here...

B) 2cos?+tan?=sec?

use

$\displaystyle tan(\alpha) \, = \, \frac{sin(\alpha)}{cos(\alpha)}$

and

$\displaystyle sec(\alpha) \, = \, \frac{1}{cos(\alpha)}$

continue from here...

 

[Deleted member 4993]

Duplicate post:

viewtopic.php?f=10&t=29635&p=113521#p113521

 

[kelleynicole30]

Thank you so much for helping me understand this a whole lot better. I just couldn't figure out what to do but I realized that I needed to square both sides first.

 

[soroban]

Hello, kelleynicole30!

$\displaystyle \text{Solve in the interval }\,[0,\,2\pi)$

. . $\displaystyle B)\;\;2\cos\alpha + \tan\alpha \;=\;\sec\alpha$

$\displaystyle \text{We have: }\;2\cos\alpha + \frac{\sin\alpha}{\cos\alpha} \;=\;\frac{1}{\cos\alpha}$

$\displaystyle \text{Multiply by }\cos\alpha\!:\;\;2\cos^2\!\alpha + \sin\alpha \;=\;1$

$\displaystyle \text{Then we have: }\;2(1-\sin^2\!\alpha) + \sin\alpha \;=\;1$

. . $\displaystyle \text{which simplifies to: }\;2\sin^2\!\alpha - \sin\alpha - 1 \;=\;0$

. . $\displaystyle \text{which factors: }\;(\sin\alpha-1)(2\sin\alpha + 1) \;=\;0$

\(\displaystyle \text{and has roots: }\;\begin{array}{ccccccc}\sin\alpha - 1 \:=\:0 & \Rightarrow &\sin\alpha \:=\:1 & \Rightarrow & \alpha \:=\:\frac{\pi}{2} & \text{extraneous root} \\ \\[-3mm] 2\sin\alpha + 1 \:=\:0 & \Rightarrow &\sin\alpha \:=\:-\frac{1}{2} & \Rightarrow & \boxed{\alpha \:=\:\frac{7\pi}{2},\:\frac{11\pi}{2}} \end{array}\)