Show that...

**Lonehwolf** · March 22nd 2009, 07:06 AM

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.

**Reckoner** · March 22nd 2009, 07:38 AM

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.

**Lonehwolf** · March 22nd 2009, 09:07 AM

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?

**Moo** · March 22nd 2009, 09:16 AM

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 ?

$\begin{aligned}<br /> 196+26 \sin(x) +\sin^2(x)<br /> &=(15 \cos(x))^2 \\<br /> &=225 \cos^2(x) \\<br /> &=225-225\sin^2(x) \end{aligned}$

and this gives your quadratic

**Reckoner** · March 22nd 2009, 09:19 AM

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

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

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

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

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

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

There is your quadratic, now solve it.

Originally Posted by Moo

Why the red things ?

$\color{red}196\color{black}+26 \sin(x) +\sin^2(x)=\dots$