solve the following trignometric equation a]sinФ+tanФ-sin2Ф=0
Answers
Answer:
okay here is your answer check below attached file and do follow me mark as brainlist answer thank yo my answer
Step-by-step explanation:
We can solve this 2nd order ODE by substituting for the variable x.
(1+x²)² y'' + 2 x (1+x²) y' + 4 y = 0 ---- (1)
Let x = tan Ф, dФ/dx = cos²Ф , x = 0, when Ф = 0.
dy/dx = dy/dФ * dФ/dx = dy/dФ * cos²Ф ---(2)
d²y/dx² = d/dФ (dy/dФ * cos²Ф) * dФ/dx
= [ cos²Ф * d²y/dФ² - 2 cosФ sinФ* dy/dФ] * cos²Ф
= cos⁴Ф * d²y/dФ² - 2 cos³Ф sinФ * dy/dФ ---(3)
So (1) becomes:
d²y/dФ² - 2 tanФ dy/dФ + 2 tanФ sec²Ф *dy/dФ * cos²Ф + 4 y = 0
d²y/dФ² = - 4 y --- (4)
This is the equation of motion for SHM... with ω = √4 = 2
Let y = A sin ωФ = A Sin2Ф , Assuming Ф = 0 = Ф₀ for y = 0
If the initial conditions are different we can have : y = A sin (Ф+Ф₀).
x = tanФ So cos2Ф = (1-x²)/(1+x²), sin2Ф = 2x/(1+x²)
Solution: y = 2 A x /(1+x²) ---- (4)
We can also verify the solution by differentiating the above :
y' = (1-x²)/(1+x²) . y'' = (x² - 3) 2x /(1+x²)³
Read more on Brainly.in - https://brainly.in/question/1111570#readmore
Answer:
Step-by-step explanation:so sinФ+sinФ/cosФ - 2sinФcosФ
by LCM,
sinФcosФ + sinФ- 2sinФcosФ/cosФ
sinФ-sinФcosФ/cosФ
sinФ(1-cosФ)/cosФ
therefore, we can write as tanФ(1-cosФ)