# Thread: Some question that need help.

1. ## Some question that need help.

1)Without evaluating the angles A and B,find the value of tan(A+2B),
given tanA = 2/3 and tanB =1/3.

2)Simplify (x/y)powerof2/3 . xpowerof3/ypowerof4

3)Find all angles between 0deg and 360deg inclusive which satisfy the equation : 3cossquartA - 2cosA - 1 = 0

4)find the value of x when : log(base2)x + log(base8)x = 4/3

BIG thanks!!!

2. 1.

$\tan{(a+b)}=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$

3. 2.

$\left(\frac{x}{y}\right) ^{\frac{2}{3}} \frac{x^3}{y^4}$

$\frac{x^{2/3}}{y^{2/3}}$ $\frac{x^3}{y^4}$

$\left(\frac{x^{11}}{y^{14}}\right)^{\frac{1}{3}}$

4. Originally Posted by Glaysher
1.

$\tan{(a+b)}=\frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$
You'd also need this trig identity seeing it is $tan(A+2B)$

$tan 2B= \frac{2tanB}{1-\tan^2B}$

obviously you can sub that into the other identity given to give:

$\tan{(a+2b)}=\frac{\tan{a}+\frac{2tanB}{1-\tan^2B}}{1-\tan{a}\frac{2tanB}{1-\tan^2B}}$

5. 4.

$\log_2{x} + \log_8{x} =\frac{4}{3}$

$\log_2{x} + \frac{\log_2{x}}{\log_2{8}} =\frac{4}{3}$

$\log_2{x} + \frac{\log_2{x}}{3} =\frac{4}{3}$

$\frac{4}{3}\log_2{x} =\frac{4}{3}$

$\log_2{x} = 1$

$x=2$

6. 3.

$3 \cos^2{A} - 2 \cos{A} - 1 = 0$

$(3 \cos {A} + 1)(\cos{A} - 1)=0$

So either $\cos{A} = -\frac{1}{3}$ or $\cos{A}=1$

Solve each equation seperately

7. thx alot...

\log_2{x} + \log_8{x} =\frac{4}{3}

\log_2{x} + \frac{\log_2{x}}{\log_2{8}} =\frac{4}{3}

\log_2{x} + \frac{\log_2{x}}{3} =\frac{4}{3}

\frac{4}{3}\log_2{x} =\frac{4}{3}---->how do you derive from this?care explain further?

\log_2{x} = 1

x=2

3 \cos^2{A} - 2 \cos{A} - 1 = 0

(3 \cos {A} + 1)(\cos{A} - 1)=0

So either \cos{A} = -\frac{1}{3} or \cos{A}=1

yeah i got 109.47deg,289.47deg,70.53 / 0deg,360
am i right ? all anges???

8. Hello, watsonmath!

Glaysher's solution to #4 is absolutely correct.

I was doing similar problems before learning the Base-change Formula.
Because of the 'related' bases (2 and 8), we were taught to manipulate the logs.
So here's my baby-talk version.

4) Solve for $x:\;\;\log_2x + \log_8x \:=\:\frac{4}{3}$

Let $\log_8x = a$. .Then: $x = 8^a$

Take logs (base 2):
. . $\log_2x \:= \:\log_2(8^a) \:= \:\log_2(2^3)^a \:=\:\log_2(2)^{3a} \:=$ $\:3a\log_22 \:=\:3a\cdot1\:=\:3a$

Hence, we have: . $\log_2x\:=\:3a\:=\:3\log_8x$

The equation becomes: . $3\log_8x + \log_8x\:=\:\frac{4}{3}\quad\Rightarrow\quad4\log_ 8x \,=$ $\frac{4}{3}\quad\Rightarrow\quad \log_8x \,= \,\frac{1}{3}
$

$\text{Therefore: }\;x \,= \,8^{\frac{1}{3}}\quad\Rightarrow\quad \boxed{x \,= \,2}$

9. Originally Posted by watsonmath

\frac{4}{3}\log_2{x} =\frac{4}{3}---->how do you derive from this?care explain further?
$\log_2 x+\frac{1}{3}\log_2 x$
$\frac{3}{3}\log_2 x+\frac{1}{3}\log_2 x$
Thus,
$\frac{4}{3}\log_2 x$

10. thx... what about the trigo qns? any helper?

11. Originally Posted by watsonmath
thx... what about the trigo qns? any helper?

12. Originally Posted by ThePerfectHacker
here
Find all angles between 0deg and 360deg inclusive which satisfy the equation : 3cossquartA - 2cosA - 1 = 0

i got 109.47deg,289.47deg,70.53 / 0deg,360
am i right ? all anges???

13. Originally Posted by watsonmath
here
Find all angles between 0deg and 360deg inclusive which satisfy the equation : 3cossquartA - 2cosA - 1 = 0

i got 109.47deg,289.47deg,70.53 / 0deg,360
am i right ? all anges???
109 is wrong. Besides for that it is good.

14. Originally Posted by ThePerfectHacker
109 is wrong. Besides for that it is good.
ic....but
A = cos-1 -1/3.... issnt that the ans for that??? which Glaysher asked me to solve???

15. Hello, watsonmath!

Find all angles $[0^o,\,360^o]$: . $3\cos^2A - 2\cos A - 1 \:= \:0$

i got 109.47°, 289.47°, 70.53°, 0°, 360°. .Am i right?

. . And your third angle is also wrong.
If $\cos A$is negative, then $A$must be in Quadrant 2 or Quadrant 3.

The expression factors: . $(3\cos A + 1)(\cos A - 1) \:=\:0$

And we have two equations to solve:

. . $3\cos A + 1 \:= \:0\quad\Rightarrow\quad \cos A \:=$ $\:-\frac{1}{3}\quad\Rightarrow\quad\boxed{A \:=\:109.46^o,\;250.53^o}$

. . $\cos A - 1 \:=\:0\quad\Rightarrow\quad \cos A = 1\quad\Rightarrow\quad\boxed{A\:=\:0^o,\;360^o}$