I tried the standard approach: sin(2x) = 2sin(x)cos(x) followed by sin(x) = sqrt(1 - cos^2(x)) and ended up with a quartic in cos(x). It does not factor. As shown by Faizan above, graphical or iterative methods are you only hope.
There it is in all it's glory. Newton's Method has the advantage of converging quickly to the root. The only tricky part is finding a good starting value for x. For example, if there is a relative max or min inside the interval for the sign change then it can easily go south and blow up. Having said that, in today's world with graphing software readily available, it is easy and fool-proof to find an x0 that will quickly converge. Well done Faizan.
For those that are interested, the recursive formula in Newton's method finds the equation of the tangent at x = x0 and then finds the x-intercept of that tangent to produce x1 - a better approximation to the root. Then just rinse and repeat to produce x2, x3 ...
You are missing an equal sign. Is it "= 0" ?
Yes, It does equal to 0.
I want the solution algebrically not graphically
We can clearly see the values of x from graph.
I tried the standard approach: sin(2x) = 2sin(x)cos(x) followed by sin(x) = sqrt(1 - cos^2(x)) and ended up with a quartic in cos(x). It does not factor. As shown by Faizan above, graphical or iterative methods are you only hope.
There it is in all it's glory. Newton's Method has the advantage of converging quickly to the root. The only tricky part is finding a good starting value for x. For example, if there is a relative max or min inside the interval for the sign change then it can easily go south and blow up. Having said that, in today's world with graphing software readily available, it is easy and fool-proof to find an x0 that will quickly converge. Well done Faizan.
For those that are interested, the recursive formula in Newton's method finds the equation of the tangent at x = x0 and then finds the x-intercept of that tangent to produce x1 - a better approximation to the root. Then just rinse and repeat to produce x2, x3 ...
Here, Newton Raphson Method is working so you may see the solution and comment.
I know about Newton's method; however, my teacher wants me to solve it algerically, neither graphically nor with newton's.
I have found out how to solve it:
Step 1: Use the trig identity of sin(2x)=2sin(x)cos(x) - https://www.quora.com/What-are-the-formulas-of-sin-2x
Step 2: Let y = tan(x/2)
Step 3: Replace sin(x) and cos(x) with half angle tan identities (using y) - https://en.wikipedia.org/wiki/Tangent_half-angle_formula
Step 4: Simplify into Quartic form - https://en.wikipedia.org/wiki/Quartic_function
Step 5: Use the Quartic formula to solve for y - https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots (either formula in full or another method)
Step 6: x = 2*atan(y)
It will look pretty horrific but it does the job and can easily be modified slightly to get all of the complex solutions