How o you solve 4sinxcosx=1 ???
There's a hidden trigonometric identity.... 2sinxcosx ==>
4sinxcosx = 1
2(2sinxcosx) = 1
2sin2x = 1
Then you solve for x :)
To solve the equation 4sin(x)cos(x) = 1, we can use some basic trigonometric identities and techniques.
Step 1: Simplify the equation using a trigonometric identity.
Multiply both sides of the equation by 2, resulting in 8sin(x)cos(x) = 2.
Step 2: Apply the double angle formula for sine.
Using the double angle formula for sine, sin(2x) = 2sin(x)cos(x), we can rewrite the equation as 4sin(2x) = 2.
Step 3: Isolate sin(2x).
Divide both sides of the equation by 4 to obtain sin(2x) = 1/2.
Step 4: Find the solutions for sin(2x) = 1/2.
We need to determine the angles whose sine value is 1/2. From the unit circle or the trigonometric table, we know that the angles whose sine is 1/2 are 30 degrees and 150 degrees (or π/6 and 5π/6 in radians).
Step 5: Solve for x.
Since we have sin(2x) = 1/2, we can set up two separate equations: 2x = 30 degrees and 2x = 150 degrees.
For the first equation, 2x = 30 degrees, we can solve for x by dividing both sides by 2: x = 15 degrees (or π/12 radians).
For the second equation, 2x = 150 degrees, we can solve for x by dividing both sides by 2: x = 75 degrees (or 5π/12 radians).
So, the solutions to the equation 4sin(x)cos(x) = 1 are x = 15 degrees (or π/12 radians) and x = 75 degrees (or 5π/12 radians).