Solve the following equation on the interval 0<x<2pi in radians or 0degrees< x <360 in degrees
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Please help me solve, 2sin^2(theta) -sin(theta) -1 = 0?
Solve the following equation on the interval 0<x<2pi in radians or 0degrees< x <360 in degrees
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Verified answer
factor
[2sin(theta)+1][sin(theta)-1]=0
set both factors equal to zero (zero product property)
2sin(theta)+1=0
sin(theta)=-1/2
sin(theta)-1=0
sin(theta)=1
theta = 90, 210, 330
To make this problem easy on yourself, replace sin (theta) with x. The problem then becomes:
2sin^2(theta) -sin(theta) -1 = 0
=> 2x^2 - x -1 = 0
Factor the equation like you would normally:
=> (2x+1)(x - 1) = 0
So now you have two answers for x:
2x+1 = 0
=> x = -1/2
and:
x-1 = 0
=> x = 1
So now replace x with sin(theta):
sin (theta) = -1/2
and
sin (theta) = 1
So because theta cannot be outside the interval 0<x<2pi, each equation will only yield one answer for theta:
theta = sin^-1(-1/2)
= 5pi/6
and
theta = sin-1(1)
= pi/2
Well, the equation itself is rather hard. Okay lets break this down into 17,0000000000002 steps
Step 1. GET A LIFE
Step 2. Google it it has an awesome calculator
Step 3. Just skip the problem its not worth it
QUESTION: DO YOU ENJOY ROADKILL?
2*sin²(t) - sin(t) - 1 = 0
2*sin²(t) - 2*sin(t) + sin(t) - 1 = 0
(2*sin(t) + 1)(sin(t) - 1) = 0
sin(t) = -1/2 and sin(t) = 1
t = 7pi/6, 11pi/6, pi/2