The-number-of-solutions-of-the-equation-sin-cos-1-sin-cos-in-the-interval-0-4pi-is-

Question Number 18457 by Tinkutara last updated on 21/Jul/17

The number of solutions of the equation

\sin θ + \cos θ = 1 + \sin θ \cos θ in the

interval [0, 4 π] is

Answered by mrW1 last updated on 21/Jul/17

\sin θ + \cos θ = 1 + \sin θ \cos θ

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

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

1 - 2 (\sin θ + \cos θ) + {(\sin θ + \cos θ)}^{2} = 0

{[(\sin θ + \cos θ) - 1]}^{2} = 0

\sin θ + \cos θ = 1

\sin θ \cos \frac{π}{4} + \cos θ \sin \frac{π}{4} = \frac{\sqrt{2}}{2}

\sin (θ + \frac{π}{4}) = \frac{\sqrt{2}}{2}

θ + \frac{π}{4} = 2 k π + \frac{π}{4}, 2 k π + \frac{3 π}{4}

θ = 2 k π, 2 k π + \frac{π}{2}

in [0, 4 π] :

θ = 0, \frac{π}{2}, 2 π, \frac{5 π}{2}, 4 π

\Rightarrow 5 solutions

Commented by mrW1 last updated on 22/Jul/17

be careful with squaring!

x = y \Rightarrow x^{2} = y^{2}

but x^{2} = y^{2} ⇏ x = y alone

since x^{2} = y^{2} \Rightarrow ∣ x ∣=∣ y ∣ i . e . x = \pm y

in your case you have also solutions

which fulfill following equation

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

e . g . with θ = π

\sin π + \cos π = - 1

1 + \sin π \cos π = 1

1 \neq - 1 but {(1)}^{2} = {(- 1)}^{2}

Commented by Tinkutara last updated on 22/Jul/17

Thanks Sir! But what is the mistake in

my method ?

\sin θ + \cos θ = 1 + \sin θ \cos θ

1 + \sin 2 θ = 1 + \sin^{2} θ \cos^{2} θ + \sin 2 θ

\sin θ \cos θ = 0 \Rightarrow \sin 2 θ = 0

θ = 0, \frac{π}{2}, π, \frac{3 π}{2}, 2 π, \frac{5 π}{2}, 3 π, \frac{7 π}{2}, 4 π

Commented by Tinkutara last updated on 22/Jul/17

Thanks Sir!