solve using identities for all angles (dont use multiple angle method)show work please cuz i am so confused.
2sin2x-3sinx=0
dunno what you mean by "dont use multiple angle method" since
sin2x = 2 sinx cosx
4sinx cosx - 3sinx = 0
sinx(4cosx - 3) = 0
sinx = 0, so x = 0,180,360,...
cosx = 3/4, so x = 41.4,318.6,...
To solve the equation 2sin(2x) - 3sin(x) = 0 using trigonometric identities, we need to use the double angle and sum/difference identities.
Let's begin by rewriting sin(2x) in terms of sin(x) using the double angle identity:
sin(2x) = 2sin(x)cos(x)
Substituting this into the equation, we get:
2(2sin(x)cos(x)) - 3sin(x) = 0
Expanding the expression:
4sin(x)cos(x) - 3sin(x) = 0
Now, we can factor out sin(x) from the equation:
sin(x)(4cos(x) - 3) = 0
So, either sin(x) = 0 or 4cos(x) - 3 = 0.
If sin(x) = 0, it means x is an integer multiple of π. Hence, one solution is x = nπ, where n is an integer.
To solve the equation 4cos(x) - 3 = 0, we isolate cos(x):
4cos(x) = 3
cos(x) = 3/4
Using the inverse cosine function, we find the principal value of x:
x = arccos(3/4)
Now, using a calculator, find the value of arccos(3/4) in radians or degrees.
To find other solutions, we can use the periodicity of the cosine function. Since the cosine function repeats every 2π, we can add 2πn to the principal value, where n is an integer.
Therefore, the final solutions to the equation 2sin(2x) - 3sin(x) = 0 using trigonometric identities for all angles are:
x = nπ, where n is an integer.
x = arccos(3/4) + 2πn, where n is an integer.