# please help me solve: cos (x - 20 degrees) = -0.437, for the range 0 to 360 degrees

• January 13th 2010, 07:41 AM
rednaxela
please help me solve: cos (x - 20 degrees) = -0.437, for the range 0 to 360 degrees
hey i'm doing some revision for a test i have tommorw and this was one of the questions that came up, i'm fine with solving other trigonometric equations but im not too sure where to go with this one eg should i replace the (x-20 degrees) for say just y and come back and then find out x, but i thought i'd ask you intelligent people just to be sure
NOTE: I HAVEN'T WORKED IN RADIANS BEFORE SO COULD WE KEEP IT IN DEGREES please =)
• January 13th 2010, 07:55 AM
General
Do you know the inverse trigonometric functions?
• January 13th 2010, 08:01 AM
rednaxela
all i know is that
cos^2(x) + sin^2(x) = 1
and that
cos x / sin x = tan x
• January 13th 2010, 08:51 AM
skeeter
Quote:

Originally Posted by rednaxela
all i know is that
cos^2(x) + sin^2(x) = 1
and that
cos x / sin x = tan x

that's not what the General asked you.

are you familiar with the inverse trig function capability of your calculator, which is what you need to solve this problem for x?

I recommend you visit this website to learn more about it ...

7. The Inverse Trigonometric Functions

also, for future reference ... $\textcolor{red}{\frac{\sin{x}}{\cos{x}} = \tan{x}}$
• January 13th 2010, 09:47 AM
rednaxela
apologies, im all over the place, yes i am aware of it
• January 13th 2010, 11:50 AM
skeeter
$
\cos(x-20) = -0.437
$

first, realize that if cosine of an angle is negative, then there are two valid angles that will work, one in quad II and one in quad III

also note that the calculator will only give you back the quad II value

$
\arccos[\cos(x-20)] = \arccos(-0.437)
$

for the quad II value of $\arccos(-0.437)$ ...

$
x-20 = \arccos(-0.437)
$

$
x = \arccos(-0.437) + 20 \approx 136^\circ
$

for the quad III value ...

$
x-20 = 360 - \arccos(-0.437)
$

$
x = 380 - \arccos(-0.437) \approx 264^\circ
$