1. ## Trigonometric Equation Help

For Negative Pi Less Than or Equal to Theta which is less than or equal to Pi
find all solutions

2(sin^2 x - cos^2 x ) = 8sinx-5

x is theta, and it is not in the exponent.

I have issues solving trigonometric equations of this style because I seem to be unsure how to handle the exponents.

I seek assistance here as usual because the lack of my educator's teaching skills.

2. Originally Posted by jd3005
For Negative Pi Less Than or Equal to Theta which is less than or equal to Pi
find all solutions

2(sin^2 x - cos^2 x ) = 8sinx-5

x is theta, and it is not in the exponent.

I have issues solving trigonometric equations of this style because I seem to be unsure how to handle the exponents.

I seek assistance here as usual because the lack of my educator's teaching skills.
$\displaystyle \cos^2 x = 1 - \sin^2 x$

so

$\displaystyle 2(sin^2 x - cos^2 x ) = 8sinx-5$

$\displaystyle 2(\sin^2 x - (1 - \sin^2 x )) = 8 \sin x - 5$

$\displaystyle 4\sin^2 x -8 \sin x +4 = 0$

$\displaystyle \sin^2 x -2 \sin x + 1 = 0$

let $\displaystyle t = \sin x$

3. you need to first use the identity:

$\displaystyle sin^2(x)-cos^2(x) = cos(2x)$

then the equation you get after this, use the identity:

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

after that you will receive a square equation with

$\displaystyle \alpha sin^2(x) + \beta sin(x) + \gamma = 0$

and you need to solve that equation by calling $\displaystyle sin(x) = t$

and $\displaystyle sin^2(x) = t^2$.

from the answers you'll get, you need to figure which one is suitable for

the condition of the question.