solve this trigonometric equation: [(2sinx+1)/(cosx)]*(cosx-sinx)+2=0

spartas

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Solve this trigonometric equation:

[(2sinx+1)/(cosx)]*(cosx-sinx)+2=0
 
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[(2sinx+1)/(cosx)]*(cosx-sinx)+2=0

What did you mean by "solve this trigonometric equation"?

Are you trying to calculate the value of "x", from the given equation?

Please share your work.
 
i have to replace the sinx with sinx=2t/1+t^2 cosx=1-t^2/1+t^2
now it looks like
\[\frac{2\frac{2t}{1+t^{2}}+1}{\frac{1-t^{2}}{1+t^{2}}}\times(\frac{1-t^{2}}{1+t^{2}}-\frac{2t}{1+t^{2}})+2=0\]
 
i have to replace the sinx with sinx=2t/1+t^2 cosx=1-t^2/1+t^2
In future, kindly please include all of the instructions when you post the exercise. Thank you!

now it looks like
\[\frac{2\frac{2t}{1+t^{2}}+1}{\frac{1-t^{2}}{1+t^{2}}}\times(\frac{1-t^{2}}{1+t^{2}}-\frac{2t}{1+t^{2}})+2=0\]
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:

. . . . .\(\displaystyle \dfrac{\left(2\frac{2t}{1\, +\,t^{2}}\,+\,1\right)}{\left(\frac{1\,-\,t^{2}}{1\,+\,t^{2}}\right)}\, \times\, \left(\dfrac{1\,-\,t^{2}}{1\,+\,t^{2}}\,-\,\dfrac{2t}{1\,+\,t^{2}}\right)\,+\,2\,=\,0\)

. . .combining w/ common denominators:

. . . . .\(\displaystyle \dfrac{\left(\frac{4t\, +\, (1\, +\, t^2)}{1\, +\, t^2}\right)}{\left(\frac{1\, -\, t^2}{1\, +\, t^2}\right)}\, \times\, \left(\dfrac{1\, -\, t^2\, -\, 2t}{1\, +\, t^2}\right)\, +\, 2\, =\, 0\)

. . .doing some simplification:

. . . . .\(\displaystyle \left(\dfrac{4t\, +\, 1\, +\, t^2}{1\, -\, t^2}\right)\, \times\, \left(\dfrac{1\, -\, 2t\, -\, t^2}{1\, +\, t^2}\right)\, +\, 2\, =\, 0\)

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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