# Thread: Help with evaluating trigonometric equations...

1. ## Help with evaluating trigonometric equations...

I am self-studying A-levels pure maths and I need a little bit of help with understanding these questions.

Find, to 1 decimnal place, all values of z in the interval -180 <= z <= 180 satisfying:

(a) coz z deg (1 + sin z deg) = 0

In this questions, it's easy to guess that it must be at -90 or 90 but I don't understand how to solve the equation.

(b) (1 - tan z deg) sin z deg = 0

I am clueless on this one.

Ideas please? Perhaps could you help me find a place that better explains this?

2. Hello, struck!

You forgot the most basic rule for solving equations . . .

. . If $a\cdot b \:=\:0$, then $a = 0$ or $b = 0.$

If a product equals zero, then one of the factors must be zero.

Find, to 1 decimal place, all values of $z$ in $[\text{-}180^o,\:180^o]$ satisfying:

$(a)\;\;\cos z(1 + \sin z) \:=\: 0$
So we have:

. . $\cos z \:=\:0 \quad\Rightarrow\quad\boxed{ z \;=\;\text{-}90^o,\:90^o}$

. . $1 + \sin z \:=\:0 \quad\Rightarrow\quad \sin z \:=\:-1 \quad\Rightarrow\quad\boxed{ z \:=\:\text{-}90^o}$

$(b)\;\;(1 - \tan z)\sin z \:= \:0$
So we have:

. . $1-\tan z \:=\:0 \quad\Rightarrow\quad \tan z \:=\:1 \quad\Rightarrow\quad\boxed{ z \;=\;45^o,\:\text{-}135^o}$

. . $\sin z \:=\:0 \quad\Rightarrow\quad\boxed{ z \;=\;\text{-}180^o,\: 0^o,\:180^o}$