find the value of x if sin2x+3cosx=2 then cos3x+sec3x is
sin2x+3cosx = 2
2sinxcosx+3cosx = 2
cosx(2sinx+3) = 2
2sinx+3 = secx
cos3x = cos2xcosx-sin2xsinx
= cos2xcosx-(2-3cosx)(2-3cosx)/(2cosx)
= (2cos^2x-1)cosx-(2-3cosx)^2/(2cosx)
I don't see where this is going. I fear that I am missing something here if you expect some simple answer.
To find the value of cos3x + sec3x, we first need to find the value of x that satisfies the equation sin2x + 3cosx = 2.
Let's break down the problem into steps:
Step 1: Rearrange the given equation sin2x + 3cosx = 2 to solve for sin2x.
sin2x = 2 - 3cosx.
Step 2: Recall the identity sin2x = 2sinxcosx.
Using the identity, we can rewrite the equation sin2x = 2 - 3cosx:
2sinxcosx = 2 - 3cosx.
Step 3: Move all terms to one side of the equation:
2sinxcosx + 3cosx - 2 = 0.
Step 4: Factor out cosx:
cosx(2sinx + 3) - 2 = 0.
Step 5: Solve for cosx:
cosx = 2 / (2sinx + 3).
At this point, we don't have enough information to directly solve for the value of x. However, we can make some observations to narrow down the available value for x.
First, notice that the denominator of cosx cannot be zero.
2sinx + 3 ≠ 0.
Second, cosx cannot be greater than 1 or less than -1.
-1 ≤ cosx ≤ 1.
Using these observations, we can conclude that:
-1 ≤ cosx ≤ 1
2sinx + 3 ≠ 0.
To find the value of cos3x + sec3x, we need to find the value of x that satisfies the equation sin2x + 3cosx = 2. Unfortunately, without additional constraints or information, we cannot determine the exact value of x or the value of cos3x + sec3x.