Q: Find all solutions to the equation in the interval [0, 2pi): sin(2x) – tan(x) = 0
A: To start, I notice we have a sin(2x) and a tan(x)… I want to get rid of the (2x) so we are just in the “x” world. To do this I need a trig identity (the double angle identity):
sin(2x) = 2sin(x)cos(x)
So, now I am going to substitute the above identity into our problem:
sin(2x) – tan(x) = 0
2sin(x)cos(x) – tan(x) = 0
Now, I know that tan(x) = sin(x) / cos(x)… Sub that in:
2sin(x)cos(x) – sin(x)/cos(x) = 0
Multiply everything by cos(x) to clear the denominator:
2 sin(x)cos²(x) – sin(x) = 0
Factor out a sin(x)
sin(x) [2cos²(x) – 1] = 0
So, either sin(x) = 0 or 2cos²(x) – 1 = 0
When does sin(x) = 0? When x = 0, π (In degrees: 0, 180)
When does 2cos²(x) – 1 = 0? Solve like so:
2cos²(x) – 1 = 0
2cos²(x) = 1
cos²(x) = 1/2
cos(x) =(+/-) √(1/2) = (+/-)1/√(2) = (+/-)√(2) / 2
When does cos(x) = √(2) / 2? When x = π/4, 7π/4 (In degrees: 45, 315)
When does cos(x) = – √(2) / 2? When x = 3π/4, 5π/4 (In degrees: 135, 225)
So, sin(2x) – tan(x) = 0 when x =0, π/4, π, 7π/4, 3π/4, 5π/4
hey, just letting you know that you missed two solutions from the cos^2(x)-1 bit, as cosx could be = 1/root2 as well as -1/root2 , while you only considered solutions for 1/root2. this is because it is a cos^2(x) so x could be (-)ve or (+)ve.
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Thank you! I just fixed this.
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