# Linear Trig Equation help plz!

• Nov 15th 2007, 07:08 AM
finalfantasy
Linear Trig Equation help plz!
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!
• Nov 15th 2007, 07:38 AM
james_bond
a)
$\displaystyle \cot\alpha=\tan\left(90^\circ-\alpha\right)$
$\displaystyle 90^\circ-x=\arctan 2.3 \approx 66.5^\circ + k\cdot 180^\circ \rightarrow x\approx 23.5^\circ + k\cdot 180^\circ$ ($\displaystyle \forall k \in \mathbb{Z}$)

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

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

Please correct me if I'm wrong!
• Nov 15th 2007, 06:57 PM
topsquark
Quote:

Originally Posted by finalfantasy
a) cot x = 2.3

Quote:

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

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

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

-Dan