# My question is 2sin^2x-2sinx-1=0 I worked through the problem and I got sin =-√3/2 But the answer in the back of the book is in decimal format and I'm not sure how to find it. It does want my to...

My question is 2sin^2x-2sinx-1=0 I worked through the problem and I got sin =-√3/2 But the answer in the back of the book is in decimal format and I'm not sure how to find it. It does want my to use the calculator. It is a problem out of the 6th edition of my Trig book, Section 6.1 problem 33. Please help me figure out how to get the correct answer.

thilina-g | College Teacher | (Level 1) Educator

Posted on

i have done a small mistake there, x should be minus,

`x = -21.5 degrees`

The general solution for sine is

`x = 180n+(-1)^n (-21.5)`

here n is any integer,

n = 0 you get, `x = -21.5`

n = 1, you get `x = 180+21.5 = 201.5`

n =2, you get `x = 360 - 21.5 = 338.5`

So the answers in 0<x<360 are 201.5 and 338.5

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

As above, the quadratic yields the solution of `sinx=(1-sqrt(3))/2`

`sin^(-1)((1-sqrt(3))/2)~~-21.5^@`

For x between 0 and 360 degrees the angles will be :

`~~180+21.5=201.5` or `~~360-21.5=338.5`

(The reference angle is 21.5 degrees, and sin is negative in the third and fourth quadrant)

spo003 | Student, College Freshman | eNotes Newbie

Posted on

The answer in the back of the book is (201.5, 338.5)

The question states,
"Use the quadratic formla to find all degree solutions and x if 0< x < 360. Use a calculator to approximate answer to the nearest tenth of a degree."

Sorry, I should have provided this before.