Solving Equations Using Trigonometric Identities

AI Thread Summary
The discussion focuses on solving the equation 2sin(x) = 3 + 2csc(x) over the interval 0 ≤ x < 2π. The user expresses confusion about how to proceed after manipulating the equation. A suggestion is made to multiply both sides by sin(x) to eliminate the csc(x) term, which leads to a clearer path towards factoring and solving for x. The user acknowledges this guidance and expresses gratitude for the clarification. The conversation emphasizes the importance of using trigonometric identities and algebraic manipulation in solving such equations.
trulyfalse
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Hey PF!

Homework Statement


Find exact solutions for the following equations over the domain 0 ≤ x <2π
2sinx = 3 + 2cscx

Homework Equations


sin2+cos2=1

The Attempt at a Solution


2sinx = 3 + 2cscx
2sinx = 3 +2(1/sinx)
sinx = 3/2 + 1/sinx
sinx - 1/sinx = 3/2
(1-1-cos2x)/sinx = 3/2
-cos2x/sinx = 3/2
cos2x/sinx = -3/2

I am perplexed by this question. Where do I go from here? How do I solve this equation?
 
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trulyfalse said:
Hey PF!

Homework Statement


Find exact solutions for the following equations over the domain 0 ≤ x <2π
2sinx = 3 + 2cscx

Homework Equations


sin2+cos2=1

The Attempt at a Solution


2sinx = 3 + 2cscx
2sinx = 3 +2(1/sinx)
sinx = 3/2 + 1/sinx
sinx - 1/sinx = 3/2
(1-1-cos2x)/sinx = 3/2
-cos2x/sinx = 3/2
cos2x/sinx = -3/2

I am perplexed by this question. Where do I go from here? How do I solve this equation?
No need to change sin2(x) into 1-cos2(x) .

Multiply both sides of sin(x) = 3/2 + 1/sin(x) by sin(x) .
 
Ahhhh... Thank you for elucidating me. I can see now that I have to factor and solve for x. Thank you again for your help!
 
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