1. ## Show that...

Show that the equation $\displaystyle 15 cos x = 13+sin x$ may be written as a quadratic equation in $\displaystyle sin x$.

Hence solve the equation, giving all values of "x" such that 0<=x<=360.

2. Originally Posted by Lonehwolf
Show that the equation $\displaystyle 15 cos x = 13+sin x$ may be written as a quadratic equation in $\displaystyle sin x$.

Hence solve the equation, giving all values of "x" such that 0<=x<=360.
Square both sides, and use the Pythagorean identity, $\displaystyle \sin^2x+\cos^2x=1.$ Collect terms and solve.

3. Didn't help me, this is what I did:

$\displaystyle 15 cos^2x-(13+sin x)^2$
$\displaystyle 225cos^4x=(13+sin x)(13+sin x)$
$\displaystyle 225cos^4x=196+26sin x +sin^2x$

Was that actually going good and I simply didn't notice or did I get what you meant with square both sides wrong?

4. Originally Posted by Lonehwolf
Didn't help me, this is what I did:

$\displaystyle {\color{red}15} cos^2x-(13+sin x)^2$
$\displaystyle 225cos^{\color{red}4}x=(13+sin x)(13+sin x)$
$\displaystyle 225cos^{\color{red}4}x=196+26sin x +sin^2x$

Was that actually going good and I simply didn't notice or did I get what you meant with square both sides wrong?
Why the red things ?

\displaystyle \begin{aligned} 196+26 \sin(x) +\sin^2(x) &=(15 \cos(x))^2 \\ &=225 \cos^2(x) \\ &=225-225\sin^2(x) \end{aligned}

5. Originally Posted by Lonehwolf
Didn't help me, this is what I did:

$\displaystyle 15 cos^2x-(13+sin x)^2$
$\displaystyle 225cos^4x=(13+sin x)(13+sin x)$
$\displaystyle 225cos^4x=196+26sin x +sin^2x$

Was that actually going good and I simply didn't notice or did I get what you meant with square both sides wrong?
I am not sure what you are doing in the above. How did squaring cosine produce $\displaystyle \cos^4x?$ Is the equation in the original post the correct equation?

You should have

$\displaystyle 15\cos x=13+\sin x$

$\displaystyle \Rightarrow225\cos^2x=\sin^2x+26\sin x+169$

$\displaystyle \Rightarrow225\left(1-\sin^2x\right)=\sin^2x+26\sin x+169$

$\displaystyle \Rightarrow225-225\sin^2x=\sin^2x+26\sin x+169$

$\displaystyle \Rightarrow-226\sin^2x-26\sin x+56=0.$

Originally Posted by Moo
Why the red things ?

$\displaystyle \color{red}196\color{black}+26 \sin(x) +\sin^2(x)=\dots$
Yes, Moo. Why the red things?

6. Ah shame on me wrong first question -.-

it should be

$\displaystyle 15cos^2x=13+sinx$

7. Originally Posted by Lonehwolf
Ah shame on me wrong first question -.-

it should be

$\displaystyle 15cos^2x=13+sinx$
Then replace $\displaystyle cos^2(x)$ with $\displaystyle 1- sin^2(x)$. This is far easier than the problem you originally posted!