# Thread:

1. ## Solving Trig Identities??

For my homework tonight, there is this one problem that is supposed to be easy but i can't figure it out.

Solve for exact solutions in the interval [0degrees, 360degrees)
[ means include, & ) means exclude
2sinx -1= cscx
I am supposed to find the degree value for x. and i think they want me to use cscx=1/sinx

2. Originally Posted by shanzwhaat
For my homework tonight, there is this one problem that is supposed to be easy but i can't figure it out.

Solve for exact solutions in the interval [0degrees, 360degrees)
[ means include, & ) means exclude
2sinx -1= cscx
I am supposed to find the degree value for x. and i think they want me to use cscx=1/sinx
They do! Then multiply by $\sin x$ and you have a quadratic equation in $\sin x$ which you can solve.

3. Thanks, but i did that and got 0degrees, 180degrees, 30 and 150
but the back of the book said the answers were 90degrees, 210degrees, and 330degrees.. Do you know what I'm doing wrong??/

4. Originally Posted by shanzwhaat
Thanks, but i did that and got 0degrees, 180degrees, 30 and 150
but the back of the book said the answers were 90degrees, 210degrees, and 330degrees.. Do you know what I'm doing wrong??/
If $2\sin x - 1 = \csc x$ then $2\sin x - 1 = \frac{1}{\sin x}$ so

$2 \sin^2 x - \sin x -1 = 0$ or $(2 \sin x + 1)(\sin x -1) = 0\;\;\Rightarrow\;\; \sin x = - \frac{1}{2}, 1$

Did you get that?

5. Originally Posted by shanzwhaat
For my homework tonight, there is this one problem that is supposed to be easy but i can't figure it out.

Solve for exact solutions in the interval [0degrees, 360degrees)
[ means include, & ) means exclude
2sinx -1= cscx
I am supposed to find the degree value for x. and i think they want me to use cscx=1/sinx
$2\sin{x} - 1 = \csc{x}$

$2\sin{x} - 1 = \frac{1}{\sin{x}}$

$2\sin^2{x} - \sin{x} = 1$

$2\sin^2{x} - \sin{x} - 1 = 0$.

Let $X = \sin{x}$ so

$2X^2 - X - 1 = 0$.

$2X^2 - 2X + X - 1 = 0$

$2X(X - 1) + 1(X - 1) = 0$

$(X - 1)(2X + 1) = 0$

$X - 1 = 0$ or $2X + 1 = 0$

$X = 1$ or $X = -\frac{1}{2}$

So $\sin{x} = 1$ or $\sin{x} = -\frac{1}{2}$.

In the domain $x \in [0^{\circ}, 360^{\circ})$, $\sin{x} = 1$ if $x = 90^{\circ}$.

In the domain $x \in [0^{\circ}, 360^{\circ}), \sin{x} = -\frac{1}{2}$ if $x = 210^{\circ}$ (Quadrant 3) or $x = 330^{\circ}$ (Quadrant 4).

Does that make sense?

6. YES! thank you everyone who answered!
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 27, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9548639059066772, "perplexity_flag": "middle"}