im not sure wheer to start and how to do solve this problem
solve: cosX(2sinX-3^1/2)=0 ,0<(or equal to)X<(or equal to)2(pie sign)
I read that as
cosx(2sinx - √3) = 0 , 0 ≤ x ≤ 2π
since it is already factored we can say
cosx=0 OR sinx = √3/2
if cosx = 0
x = π/2 or x = 3π/2 , (90° or 270°)
if sinx = √3/2
then x must be in quadrants I or II
x = π/3 or x = 2π/3 , (60° or 120°)
i didn't know how to put the root in so i converted it by adding in the 1/2 exponent! thank you :)
To solve the equation cos(X)(2sin(X) - √3) = 0, where 0 ≤ X ≤ 2π, we need to find the values of X that satisfy the equation.
To start, we can use the zero product property. According to this property, if a product of factors is equal to zero, then at least one of the factors must be zero. In other words, if any of the factors is equal to zero, then the whole expression is equal to zero.
In this case, we have two factors: cos(X) and (2sin(X) - √3). We can set each factor equal to zero and solve for X separately.
Setting cos(X) = 0:
To find the values of X for which cos(X) = 0, we can use the unit circle or the cosine function's periodicity. The unit circle shows that cosine equals zero at π/2 and 3π/2. Since the range for X is stated as 0 ≤ X ≤ 2π, we know that X = π/2 and X = 3π/2 are within the valid range.
Setting 2sin(X) - √3 = 0:
To solve this equation for sin(X), we isolate sin(X) by moving √3 to the other side:
2sin(X) = √3
sin(X) = √3/2
The value of √3/2 is a special value for sine, corresponding to π/3 in the unit circle. Therefore, we know that X = π/3 and X = 2π - π/3 = 5π/3 are valid solutions.
So, the values of X that satisfy the equation cos(X)(2sin(X) - √3) = 0, where 0 ≤ X ≤ 2π, are:
X = π/2, X = 3π/2, X = π/3, and X = 5π/3.