Math Help - trigonometry question

1. trigonometry question

solve for x , -180<x<180

2cos x+1= sin x

This is what i did ..

sin x -2 cos x=1

$\sqrt{5}\sin (x-63.43)=1$

so x-63.43=26.57

x=90

and x-63.43=-26.57 , -153.43

x=36.86 , -90

My answers are in red . But tat doesn't agree with the answer in the book . Thanks for helping .

2. Originally Posted by thereddevils
sin x -2 cos x=1

$\sqrt{5}\sin (x-63.43)=1$

so x-63.43=26.57
How did you go from the first line to the second?

3. Originally Posted by thereddevils
solve for x , -180<x<180

2cos x+1= sin x

This is what i did ..

sin x -2 cos x=1

$\sqrt{5}\sin (x-63.43)=1$

so x-63.43=26.57

x=90

and x-63.43=-26.57 , -153.43

x=36.86 , -90

My answers are in red . But tat doesn't agree with the answer in the book . Thanks for helping .
2 cos x + 1 = sin x
$2 \cos x + 1 = \sqrt{1 - \cos^2x}$

Squaring both sides, you get:

$4 \cos^2x+ 4 \cos x = 1 - \cos^2 x$

Move everything to one side:
$5\cos^2x + 4 \cos x - 1 = 0$
Sub: $u = \cos x$
$5u^2 + 4u -1 = 0$

$u = \frac{-4 \pm 6}{10}$

giving -1 or 0.2.

Set cosine x equal to each and solve. Check your solutions to make sure they work.

Good luck!!

4. Originally Posted by Jameson
How did you go from the first line to the second?
I used this r sin (x-a) , where $r=\sqrt{1^2+2^2}=\sqrt{5}$

and tan a = 2

do you see anything wrong in my working ?

5. Originally Posted by thereddevils
solve for x , -180<x<180

2cos x+1= sin x

This is what i did ..

sin x -2 cos x=1

$\sqrt{5}\sin (x-63.43)=1$

so x-63.43=26.57

x=90 Mr F says: This is correct.

and x-63.43=-26.57 , -153.43 Mr F says: This is wrong. It should be x - 63.43 = 180 - 26.57 = 153.43. Note that sin(-A) = -sin(A) .....

[snip]
Therefore x - 63.43 = 153.43 => x = 216.86 degrees which is outside the domain therefore subtract 360 degrees to get the other value of x.