Solve this trigonometric equation:
[(2sinx+1)/(cosx)]*(cosx-sinx)+2=0
Solve this trigonometric equation:
[(2sinx+1)/(cosx)]*(cosx-sinx)+2=0
Last edited by stapel; 02-01-2017 at 11:47 AM. Reason: Copying instructions from subject line back into post.
to find the x, it has to do with tangent half-angle formulas!
i have to replace the sinx with sinx=2t/1+t^2 cosx=1-t^2/1+t^2
now it looks like
\[\frac{2\frac{2t}{1+t^{2}}+1}{\frac{1-t^{2}}{1+t^{2}}}\times(\frac{1-t^{2}}{1+t^{2}}-\frac{2t}{1+t^{2}})+2=0\]
In future, kindly please include all of the instructions when you post the exercise. Thank you!
Okay, so you're left with just algebra to do. Use what you learned about complex fractions (here), solving polynomials (here), and the Quadratic Formula (here) to simplify and solve.
. . .the equation you posted:
. . . . .[tex]\dfrac{\left(2\frac{2t}{1\, +\,t^{2}}\,+\,1\right)}{\left(\frac{1\,-\,t^{2}}{1\,+\,t^{2}}\right)}\, \times\, \left(\dfrac{1\,-\,t^{2}}{1\,+\,t^{2}}\,-\,\dfrac{2t}{1\,+\,t^{2}}\right)\,+\,2\,=\,0[/tex]
. . .combining w/ common denominators:
. . . . .[tex]\dfrac{\left(\frac{4t\, +\, (1\, +\, t^2)}{1\, +\, t^2}\right)}{\left(\frac{1\, -\, t^2}{1\, +\, t^2}\right)}\, \times\, \left(\dfrac{1\, -\, t^2\, -\, 2t}{1\, +\, t^2}\right)\, +\, 2\, =\, 0[/tex]
. . .doing some simplification:
. . . . .[tex]\left(\dfrac{4t\, +\, 1\, +\, t^2}{1\, -\, t^2}\right)\, \times\, \left(\dfrac{1\, -\, 2t\, -\, t^2}{1\, +\, t^2}\right)\, +\, 2\, =\, 0[/tex]
Multiply through by the denominators to clear the fractions, and so forth. If you get stuck in finding the four solutions, please reply showing all of your steps so far, starting with the ones displayed above. Thank you!
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