1. ## Show that...

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

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

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

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

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

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

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:

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

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 ?

\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:

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

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 $\cos^4x?$ Is the equation in the original post the correct equation?

You should have

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

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

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

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

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

Originally Posted by Moo
Why the red things ?

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

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

it should be

$15cos^2x=13+sinx$

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

it should be

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