Solving Trig Equations

**Macleef** · November 24th 2007, 01:57 PM

Solve each equation for $0 \leq \theta \leq 360$

--
$sin^2x - 1 = cos^2x$
--

Work:
$cos^2x - cos^2x = 0$
$0 = 0$

--
Textbook Answers:
90 and 270
--

Incorporate the use of the double angle formulas identities and other identities to solve the following on $0 \leq \theta \leq 360$

--
$sin2x = 2cos2x$
--

Work:
$2sinxcosx - 2cosx = 0$
$2cosx (sinx - 1) = 0$

What do I do now?

--
Textbook Answers:
90 and 270
--

Btw, how would you simply $cos^2x$ to $cosx$ if you had something like $cos^2x = 1$ ? Would it be...
$cosx = \sqrt{1}$
$x = cos^-1 \sqrt{1}$ ?

**topsquark** · November 24th 2007, 02:10 PM

Originally Posted by Macleef

--
$sin^2x - 1 = cos^2x$
--

Work:
$cos^2x - cos^2x = 0$
$0 = 0$

Not quite. $1 - sin^2(x) = cos^2(x)$ , so
$sin^2x - 1 = cos^2x$

$-cos^2(x) = cos^2(x)$

$2cos^2(x) = 0$

$cos(x) = 0$

Thus $x = 90^o , 270^o$

-Dan

**topsquark** · November 24th 2007, 02:23 PM

Originally Posted by Macleef

Incorporate the use of the double angle formulas identities and other identities to solve the following on $0 \leq \theta \leq 360$

--
$sin2x = 2cos2x$
--

Work:
$2sinxcosx - 2cosx = 0$
$2cosx (sinx - 1) = 0$

What do I do now?

You forgot about the cos(2x).

And check your equation. Your given solutions don't work.

This is one of the nastier ones:
$sin(2x) = 2cos(2x)$

$2sin(x)cos(x) = 2cos(2x)$

$sin(x) cos(x) = cos(2x)$

$sin(x) cos(x) = 2cos^2(x) - 1$

$cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1$

$cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

$5cos^4(x) - 5cos^2(x) + 1 = 0$

This has solutions
$cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0$
and
$cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o$

Of these, only $x = 31.7175^o, 121.717^o$ solve the original equation.

-Dan

**topsquark** · November 24th 2007, 02:25 PM

Originally Posted by Macleef

Btw, how would you simply $cos^2x$ to $cosx$ if you had something like $cos^2x = 1$ ? Would it be...
$cosx = \sqrt{1}$
$x = cos^-1 \sqrt{1}$ ?

How do you solve any equation where you take the square root of both sides?

$cos^2(x) = 1$

$cos(x) = \pm 1$

Thus
$x = cos^{-1}(1) = 0^o$
and
$x = cos^{-1}(-1) = 180^o$

-Dan

**Macleef** · November 24th 2007, 02:59 PM

Originally Posted by topsquark

You forgot about the cos(2x).

And check your equation. Your given solutions don't work.

This is one of the nastier ones:
$sin(2x) = 2cos(2x)$

$2sin(x)cos(x) = 2cos(2x)$

$sin(x) cos(x) = cos(2x)$

$sin(x) cos(x) = 2cos^2(x) - 1$

$cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1$

$cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

$5cos^4(x) - 5cos^2(x) + 1 = 0$

This has solutions
$cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0$
and
$cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o$

Of these, only $x = 31.7175^o, 121.717^o$ solve the original equation.

-Dan

What the heck...? I'm so confused
I'm pretty sure this equation isn't that complicated...

**Macleef** · November 24th 2007, 03:01 PM

Originally Posted by topsquark

Not quite. $1 - sin^2(x) = cos^2(x)$ , so
$sin^2x - 1 = cos^2x$

$-cos^2(x) = cos^2(x)$

$2cos^2(x) = 0$

$cos(x) = 0$

Thus $x = 90^o , 270^o$

-Dan

How did you get $-cos^2x$ ? Isn't it $cos^2x$ for $sin^2x - 1$ ?

**topsquark** · November 25th 2007, 06:26 AM

Originally Posted by Macleef

What the heck...? I'm so confused
I'm pretty sure this equation isn't that complicated...

This equation is. Whatever your book (teacher, professor, whatever) intended for you to solve certainly wasn't given the listed solution set. There's a typo in your given equation, but I can't figure out what it was meant to be.

-Dan

**topsquark** · November 25th 2007, 06:27 AM

Originally Posted by Macleef

How did you get $-cos^2x$ ? Isn't it $cos^2x$ for $sin^2x - 1$ ?

$sin^2(x) + cos^2(x) = 1$

$sin^2(x) + cos^2(x) - 1 = 0$

$sin^2(x) - 1 = -cos^2(x)$

-Dan

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