**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, 7****π/4 (In degrees: 45, 315)**

When does cos(x) = – √(2) / 2? When x = **3π/4, 5****π/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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