To solve the equation log2(-3sinx) = 2log2(cosx) + 1 for x, we need to simplify the equation and use logarithmic properties.
Step 1: Simplify the equation
Using the property loga(b) = c is equivalent to a^c = b, we can rewrite the equation as follows:
2log2(cosx) = log2(-3sinx) - 1
Step 2: Apply logarithmic properties
By applying the logarithmic property loga(b) - loga(c) = loga(b/c), we can rewrite the equation as:
2log2(cosx) = log2((-3sinx)/2)
Step 3: Convert to exponential form
Rewriting the equation in exponential form, we have:
2^2log2(cosx) = 2^log2((-3sinx)/2)
Simplifying further, we get:
4(cosx) = (-3sinx) / 2
Step 4: Solve for x
To solve for x, we multiply both sides of the equation by 2:
8(cosx) = -3sinx
Expanding further:
8cosx = -3sinx
Now, we divide both sides by cosx:
8 = -3tanx
Dividing both sides by -3 gives:
-8/3 = tanx
Step 5: Find the inverse tangent
To find the value of x, we take the inverse tangent (arctan) of both sides:
x = arctan(-8/3)
The solution for x is x = arctan(-8/3).