Solving Trig Identities??

**shanzwhaat** · January 13th 2009, 02:47 PM

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
i'm not sure, i already tried and it didn't work. Please help me!!

**Danny** · January 13th 2009, 02:50 PM

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
i'm not sure, i already tried and it didn't work. Please help me!!

They do! Then multiply by $\sin x$ and you have a quadratic equation in $\sin x$ which you can solve.

**shanzwhaat** · January 13th 2009, 03:21 PM

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??/

**Danny** · January 13th 2009, 03:27 PM

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?

**Prove It** · January 13th 2009, 03:32 PM

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
i'm not sure, i already tried and it didn't work. Please help me!!

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

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

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

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

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?

**shanzwhaat** · January 13th 2009, 03:36 PM

YES! thank you everyone who answered!

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