Let-a-is-a-real-number-How-many-solutions-can-the-equation-in-sin-cos-sin-cos-1-a-have-for-0-lt-lt-pi-2-

Question Number 59113 by naka3546 last updated on 04/May/19

L e t a i s a r e a l n u m b e r . H o w m a n y s o l u t i o n s

c a n t h e e q u a t i o n i n θ

(\sin θ + \cos θ) (\sin θ \cos θ - 1) = a

h a v e f o r 0 < θ < \frac{π}{2} ?

Answered by MJS last updated on 05/May/19

(\sin θ + \cos θ) (\sin θ \cos θ - 1) = a

\sin^{2} θ \cos θ - \sin θ + \sin θ \cos^{2} θ - \cos θ = a

(1 - \cos^{2} θ) \cos θ - \sin θ + \sin θ (1 - \sin^{2} θ) - \cos θ = a

\cos θ - \cos^{3} θ - \sin θ + \sin θ - \sin^{3} θ - \cos θ = a

- \cos^{3} θ - \sin^{3} θ = a

for θ \in [0; - \frac{π}{2}] :

f (θ) = - \cos^{3} θ - \sin^{3} θ has got a local maximum

at (\begin{matrix} \frac{π}{4} \\ - \frac{\sqrt{2}}{2} \end{matrix}) and local mimima at (\begin{matrix} 0 \\ - 1 \end{matrix}) and (\begin{matrix} \frac{π}{2} \\ - 1 \end{matrix})

\Rightarrow for a > - \frac{\sqrt{2}}{2} we have 0 solutions

for a = - \frac{\sqrt{2}}{2} we have 1 solution

for - 1 ⩽ a < - \frac{\sqrt{2}}{2} we have 2 solutions

for a < - 1 we have 0 solutions

Commented by naka3546 last updated on 05/May/19

s y y = a ?

I s i t a m i s t a k e i n w r i t i n g, s i r ?

Commented by MJS last updated on 05/May/19

sorry just a weird typo . I corrected it .

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