Solutions to trig equations

**RosieLaird** · August 9th 2010, 08:44 AM

a)

b)

**eumyang** · August 9th 2010, 08:58 AM

a) $\cos x + 1 = \sin x$

Square both sides:
$(\cos x + 1)^2 = \sin^2 x$
$\cos^2 x + 2\cos x + 1 = \sin^2 x$

Use the Pythagorean identity on the RHS:
$\cos^2 x + 2\cos x + 1 = 1 - \cos^2 x$

Move everything to the LHS and you'll have a quadratic on cos x. Solve for cos x, and then find x. Can you take it from here?

b) $2\cos 3x - 1 = 0$

Solve for cos 3x:
$2\cos 3x = 1$
$\cos 3x = \frac{1}{2}$

Now ask yourself, the cosine of what angle(s) is 1/2? (You should know; they are all special angles!) The answers you come up with equal to 3x, which means that you'll have to divide each of the answers by 3 to solve for x.

**Mazerakham** · August 9th 2010, 09:26 AM

You would have to prove this mathematically, but clearly from the graph, the first equation has solutions $\pi \left( {\cfrac{3}{4}\ \pm\ \cfrac{1}{4} \right)\ +\ 2\pi n\ \ \left( n \in \mathbb Z \right)$
For the second, you need $cos(3x)=\cfrac{1}{2}$ . You should be able to do that.

**Mazerakham** · August 9th 2010, 09:39 AM

Originally Posted by eumyang

a) $\cos x + 1 = \sin x$

[...]

Move everything to the LHS and you'll have a quadratic on cos x. Solve for cos x, and then find x. Can you take it from here?

Eumyang is right, although this method does lead to an extraneous answer, so you have to be careful: we have:

$2cos^2 x + 2 cos x = 0$
$cos x \left( cos x + 1 \right) = 0$
$cos x = 0\ or\ -1$

However, this would lead to one set of solutions $\cfrac{3\pi}{2} + 2\pi n$ which actually doesn't fit the original equation. The problem arises from the fact that, given the equations

$(1)\ \ LHS \ = \ RHS$
$(2)\ \ (LHS)^2 \ = \ (RHS)^2$

we have (1) implies (2), but not (2) implies (1).

**eumyang** · August 9th 2010, 09:50 AM

Originally Posted by Mazerakham

Eumyang is right, although this method does lead to an extraneous answer, so you have to be careful:

Uhm, I was kind of hoping that RosieLaird would have realized it on her own. That's why I said, "Can you take it from here?", you know. Thanks for giving it away.

**Archie Meade** · August 9th 2010, 10:06 AM

Originally Posted by RosieLaird

a)

b)

(a)

Another way...

Let (x,y) be a point on the circumference of the unit circle.

$cos\theta+1=sin\theta$

$sin\theta-cos\theta=1$

$\displaystyle\huge\frac{y}{r}-\frac{x}{r}=1$

$r=1\ \Rightarrow\ y-x=1$

By Pythagoras' theorem

$\sqrt{x^2+y^2}=1$

$\sqrt{(y-x)^2}=\sqrt{y^2+x^2}$

$\sqrt{y^2-2xy+x^2}=\sqrt{y^2+x^2}$

$xy=0\ \Rightarrow\ x=0\ or\ y=0$

Test the 4 possible angles

$cos(0)+1=2\ \ne\ sin(0)$

$cos\left(90^o\right)+1=1=sin\left(90^o\right)$

$cos\left(180^o\right)+1=0=sin\left(180^o\right)$

$cos\left(270^o\right)+1=1\ \ne\ sin\left(270^o\right)$

**Mathelogician** · August 10th 2010, 05:30 AM

Note:
If squaring both sides as eumyang did, you must al last test the obtained x. Because it may that some soloutions are not real!

-----------------------------
Also you can use the method of solving first type of classical trig equations...

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