solve for (<_ = less than or equal to / pie = pie sign / -pie = negative pie)
3 sin²x = cos²x ; 0 <_ x < 2pie
cos²x - sin²x = sinx ; -pie < x <_ pie
1st one:
3 sin²x = cos²x ; 0 ≤ x < 2π
sin²x /cos²x = 1/3
tan^2 x = 1/3
tanx = ± 1/√3
set your calculator to radians ....
x = .5236
x = π - .5236 = 2.61799
x = π + .5236 = 2.61799
x = 2π- .5236 = 5.7596
2nd:
cos^2 x - sin^2 x = sinx
1 - sin^2 x - sin^2 x - sinx = 0
2sin^2 x + sinx - 1 = 0
(2sinx -1)(sinx + 1) = 0
sinx = 1/2 or sinx = -1
if sinx = 1/2
x = π/6, or x = 5π/6
if sinx = -1
then x = =π/2 or 3π/2 ,but -π < x ≤ π
so x = -π/2
x = -π/2 , π/6 , 5π/5
Thank you Reiny! The first question (1st one) your calculations for the 2nd x is wrong. x = x = π + .5236 = 3.665192654 and not 2.61799
yup, clearly a typo, since the first part of it is correct
To solve the first equation, we can use the trigonometric identity: sin²x + cos²x = 1.
1. Start with the given equation: 3 sin²x = cos²x
2. Rearrange the equation by subtracting cos²x from both sides: 3 sin²x - cos²x = 0
3. Apply the trigonometric identity sin²x + cos²x = 1: 3 sin²x - (1 - sin²x) = 0
4. Simplify the equation: 3 sin²x - 1 + sin²x = 0
5. Combine like terms: 4 sin²x - 1 = 0
6. Add 1 to both sides: 4 sin²x = 1
7. Divide both sides by 4: sin²x = 1/4
8. Take the square root of both sides (since we're looking for sinx): sinx = ±√(1/4)
9. Simplify: sinx = ±1/2
So, the solutions for sinx are sinx = 1/2 and sinx = -1/2.
To solve the second equation, we can simply rearrange it to isolate the variable x.
1. Start with the equation: cos²x - sin²x = sinx
2. Rearrange by subtracting sinx from both sides: cos²x - sin²x - sinx = 0
3. Since we already know the values of sinx from the first equation, substitute those values into the equation and solve for x.
For sinx = 1/2:
cos²x - (1/2)² - (1/2) = 0
cos²x - 1/4 - 1/2 = 0
cos²x - 3/4 = 0
cos²x = 3/4
cosx = ±√(3/4)
cosx = ±√3/2
For sinx = -1/2:
cos²x - (-1/2)² - (-1/2) = 0
cos²x - 1/4 + 1/2 = 0
cos²x + 1/4 = 0
cos²x = -1/4 (Since cos²x cannot be negative, this equation has no solutions in the given interval.)
Therefore, the solutions for the second equation are cosx = √3/2 and cosx = -√3/2 in the interval -π < x ≤ π.