# Thread: I can't solve these Trig Equations

1. ## I can't solve these Trig Equations

Hi , out of my 100 question holiday packet, these are ones I just couldn't understand:

Solve these equations in the indicated domain

(1+cos(theta))/(sin(theta))=-1
domain: [-180,180)

cos4(theta) - sin2(theta) = 0
domain: (-90,90)

cos4(theta) - sin2(theta) = 1
domain: [-90,90)

cos3(theta) + cos5(theta) = 0
domain: (-90,90)

sin5(theta) + sin7(theta) = 0
domain: [-90,90)

cosx - root3(sinx) = 1
domain (0,2pi]

sinx - root3(cosx) = 1
domain: [-pi,pi]

[tan(10theta) + tan50(deg)]/[1-tan(10theta)tan50(deg)] = (root3)/3
domain: (0,90)

tan(theta) - tan(10theta) = 1 + tan(theta)tan10(deg)
domain: [-180,180]

tan(1/2)x + 1 = cosx
domain: [0,4pi]

2(cos^2)(1/2)x - 2 = 2cosx
domain: [-pi,pi)

2cos(theta + 30)cos(theta - 30) = 1
domain: [-180,180]

4sin(theta + 75)cos(theta - 75) = 1
domain: [-180,180)

(cos^2)(1/2)x - (1/2)cosx = 1/2
domain: real numbers

sinxtan(1/2)x = 1 - cosx
domain: real numbers

Any help is greatly appreciated!

Thanks!

2. Hello, hassapi!

Here's #8 . . .

$\displaystyle 8)\;\;\frac{\tan 10\theta + \tan50^o}{1-(\tan 10\theta)(\tan50^o)} \:=\:\frac{\sqrt{3}}{3}\qquad\text{ domain: }(0^o,90^o)$
We're expected to recognize the left side as a Compound Angle Identity . . .

. . $\displaystyle \tan(10\theta + 50^o) \;=\;\frac{1}{\sqrt{3}}$

And we should know that the angle is 30° and its variations.

So we have:. . $\displaystyle 10\theta + 50^o \;=\;\{30^o,\:210^o,\:390^o,\:570^o,\:750^o,\:930^ o\}$

. . . . . Then: . . . . .$\displaystyle 10\theta \;=\;\{{\color{red}\rlap{/////}}-20^o,\:160^o,\:340^o,\;520^o,\:700^o,\:880^o\}$

. . Therefore: . . . . . .$\displaystyle \theta \;=\;\{16^o,\;34^o,\:52^o,\:70^o,\:88^o\}$

3. Originally Posted by Soroban
Hello, hassapi!

Here's #8 . . .

We're expected to recognize the left side as a Compound Angle Identity . . .

. . $\displaystyle \tan(10\theta + 50^o) \;=\;\frac{1}{\sqrt{3}}$

And we should know that the angle is 30° and its variations.

So we have:. . $\displaystyle 10\theta + 50^o \;=\;\{30^o,\:210^o,\:390^o,\:570^o,\:750^o,\:930^ o\}$

. . . . . Then: . . . . .$\displaystyle 10\theta \;=\;\{{\color{red}\rlap{/////}}-20^o,\:160^o,\:340^o,\;520^o,\:700^o,\:880^o\}$

. . Therefore: . . . . . .$\displaystyle \theta \;=\;\{16^o,\;34^o,\:52^o,\:70^o,\:88^o\}$

Ah, thank you very much! I don't think my teacher taught us compound angle identity...