Determine the solutions to the equation 2sinx − cos2x = sin2x for 0 ≤ x ≤ 2π .
How do I start this question?
do you mean sin(2x) or sin^2 x ???
Sorry, I meant 2sinx - cos^2x = sin^2x for 0 ≤ x ≤ 2π
2sinx - cos^2x = sin^2x
Do I change the left side to
2sinxcosx - cos^2x = sin^2x
2 sin x = sin^2 x + cos^2 x = 1
sin^2 x + cos^2 x = 1 VERY IMPORTANT trig identity
sin x = 1/2
x = 30 degrees or 150 degrees
Thank you~
May I know how you got 150 degrees? I am so confused when finding the reference angles, whether to subtract or add the degrees.
well I just looked for 30 degrees above and below the x and - x axis and picked the one where sin was positive. That is in quadrants 1 and 2. 180-30 = 150
Thank you!
To start solving the equation 2sin(x) − cos^2(x) = sin^2(x), you can follow these steps:
Step 1: Simplify the equation.
Rearrange the equation by moving all terms to one side to make it easier to solve. In this case, let's move all terms to the left side:
2sin(x) − cos^2(x) - sin^2(x) = 0
Step 2: Simplify further.
Use trigonometric identities to simplify the equation. In this case, use the identity sin^2(x) + cos^2(x) = 1:
2sin(x) − (1 - cos^2(x)) - sin^2(x) = 0
2sin(x) - 1 + cos^2(x) - sin^2(x) = 0
Step 3: Combine like terms.
Combine the terms with sin(x) and simplify the expression:
-cos^2(x) + sin^2(x) + 2sin(x) - 1 = 0
Step 4: Factor or use quadratic equation.
From Step 3, we have a quadratic-like expression: -cos^2(x) + sin^2(x) + 2sin(x) - 1. To solve for x, you can either factor the expression or use the quadratic formula.
Step 5: Solve for x.
There are different methods you could use to solve a quadratic equation; however, in this case, introducing a new variable will make the process easier. Let u = sin(x):
-sin^2(x) + sin^2(x) + 2sin(x) - 1 = 0
Canceling out sin^2(x), we get:
2u - 1 = 0
2u = 1
u = 1/2
Now, substitute back sin(x) = 1/2 and solve for x:
sin(x) = 1/2
From the unit circle or trigonometric values, you can determine that the solutions for sin(x) = 1/2 are x = π/6 and x = 5π/6.
So the solutions to 2sin(x) − cos^2(x) = sin^2(x), for 0 ≤ x ≤ 2π, are x = π/6 and x = 5π/6.