Linear Trig Equation help plz!

**finalfantasy** · November 15th 2007, 07:08 AM

2. Find all solutions of each equation.
a) cot x = 2.3 (I can find it with special angles, but not this)
b) 5 cos 3x = -3

3. c) 2sinx - 1 = -2

You people here are very kind! Thanks!

**james_bond** · November 15th 2007, 07:38 AM

a)
$\cot\alpha=\tan\left(90^\circ-\alpha\right)$
$90^\circ-x=\arctan 2.3 \approx 66.5^\circ + k\cdot 180^\circ \rightarrow x\approx 23.5^\circ + k\cdot 180^\circ$ ( $\forall k \in \mathbb{Z}$ )

b)
$x=\frac{\arccos\frac{-3}{5}}{3}\approx \frac{\pm 126.87+k\cdot 360^\circ}{3}\approx \pm 42.29 + k\cdot 120^\circ$ ( $\forall k \in \mathbb{Z}$ )

c) $2\sin x=-1$
$x=\arcsin\frac{-1}{2}=-30^\circ + k\cdot 360^\circ$ or $180^\circ -(-30^\circ) + k\cdot 360^\circ = 210^\circ + k\cdot 360^\circ$ ( $\forall k \in \mathbb{Z}$ )
Or if it should be $2\sin (x-1)=-2$ then $x-1=\arcsin-1=-90^\circ + k\cdot 360^\circ$ or $270^\circ + k\cdot 360^\circ \rightarrow x=-89^\circ + k\cdot 360^\circ$ or $269^\circ + k\cdot 360^\circ$ ( $\forall k \in \mathbb{Z}$ )

Please correct me if I'm wrong!

**topsquark** · November 15th 2007, 06:57 PM

Originally Posted by finalfantasy

a) cot x = 2.3

Originally Posted by james_bond

a)
$\cot\alpha=\tan\left(90^\circ-\alpha\right)$
$90^\circ-x=\arctan 2.3 \approx 66.5^\circ + k\cdot 180^\circ \rightarrow x\approx 23.5^\circ + k\cdot 180^\circ$ ( $\forall k \in \mathbb{Z}$ )

It is simpler to note that
$cot(x) = \frac{1}{tan(x)} = 2.3$

So
$x = tan^{-1} \left ( \frac{1}{2.3} \right )$

-Dan

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