3sin(2x)-2cos(x)=0
Use the equation
sin(2x) = 2 sin x cos x
so that
6 sinx cosx -2 cosx = 0
cosx*(1 - 3sinx) = 0
(I divided out -2)
So the equation is satisfied wherever cosx = 0 or sinx = 1/3
x = pi/2 and 0.3398 radians are two solutions. There are others in other quadrants.
To find the solutions to the equation 3sin(2x) - 2cos(x) = 0, we can use trigonometric identities. Let's break it down step by step:
Step 1: Simplify the equation using trigonometric identities.
Using the double-angle identity, sin(2x) = 2sin(x)cos(x), we can rewrite the equation as:
3 * 2sin(x)cos(x) - 2cos(x) = 0
Simplifying further, we have:
6sin(x)cos(x) - 2cos(x) = 0
Step 2: Factor out the common term.
Taking out the common factor of cos(x), we get:
cos(x) * (6sin(x) - 2) = 0
Step 3: Set each factor equal to zero.
Since a product is equal to zero only if one or more of its factors is zero, we set each factor equal to zero and solve for x.
cos(x) = 0
To find the solutions for this equation, we need to determine where the cosine function equals zero. The cosine function is equal to zero at π/2 + nπ (where n is any integer).
Thus, we have x = π/2 + nπ.
Next,
6sin(x) - 2 = 0
We solve this equation separately:
6sin(x) = 2
sin(x) = 2/6
sin(x) = 1/3
We find the reference angle by taking the inverse sine of 1/3:
Reference angle = sin^(-1)(1/3) ≈ 0.3398 radians
Since sin(x) = 1/3 is positive in both the first and second quadrants, we have:
x = 0.3398 + 2πn, and x = π - 0.3398 + 2πn, where n is any integer.
So, the solutions to the original equation are:
x = π/2 + nπ, and x = 0.3398 + 2πn, and x = π - 0.3398 + 2πn, where n is any integer.