Trigonometric equations with 2 functions Given the equation:

Malachi Mullins

Malachi Mullins

Answered question

2022-04-08

Trigonometric equations with 2 functions
Given the equation: sin2x+cosx=0

Answer & Explanation

Quimpoah3b

Quimpoah3b

Beginner2022-04-09Added 13 answers

You are on the right track. The equation
sin2x+cosx=0
becomes
cos2x+cosx+1=0
with the substitution sin2x=1cos2x. At this point, you may solve the quadratic equation in cosx to find
cosx=1±1+42=152
that is, formally,
cosx= (φ),(1φ):
where φ=1.618033 is the golden ratio. However, since cosx[1,+1] the option cosx=φ must be discarded, so that
cosx=1φ=152
One solution is
x=arccos1522.237
in radians (in degrees, about 128.2); however, since cos is an even function, -x must be a solution too. Finally, cos is 2π-periodic, therefore the other solutions may be found by adding a factor of 2πn to x and -x, with nZ

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