The-number-of-solutions-of-sin3x-cos2x-0-in-0-3pi-2-is-

Question Number 18545 by Tinkutara last updated on 24/Jul/17

The number of solutions of

\sin 3 x + \cos 2 x = 0 in [0, \frac{3 π}{2}] is

Answered by 433 last updated on 24/Jul/17

\sin (3 x) = \sin (2 x + x) = s i n (2 x) c o s (x) + s i n (x) c o s (2 x)

= 2 \sin (x) \cos^{2} (x) + \sin (x) \times ({2 \cos}^{2} (x) - 1)

4 \sin (x) \cos^{2} (x) - \sin (x) = \sin (x) ({4 \cos}^{2} (x) - 1)

\cos (2 x) = 2 \cos^{2} (x) - 1

\sin 3 x + \cos 2 x = 0 \Leftrightarrow

\sin (x) (4 \cos^{2} (x) - 1) + 2 \cos^{2} (x) - 1 = 0

\sin (x) (4 - 4 \sin^{2} (x) - 1) + 2 - 2 \sin^{2} (x) - 1 = 0

- 4 \sin^{3} (x) - 2 \sin^{2} (x) + 3 \sin (x) + 1 = 0

y = s i n (x)

4 y^{3} + 2 y^{2} - 3 y - 1 = 0

(y + 1) (4 y^{2} - 2 y - 1) = 0

y = - 1 \Rightarrow \sin (x) = - 1 \Rightarrow x = \frac{3 π}{2}

4 y^{2} - 2 y - 1 = 0

Δ = 2^{2} - 4 \times 4 \times (- 1) = 20

y = \frac{2 \pm \sqrt{20}}{8} \Rightarrow y = \frac{1 \pm \sqrt{5}}{4} \Rightarrow y = \frac{ϕ}{2} o r y = \frac{1 - ϕ}{2}

\sin (x) = \frac{ϕ}{2} \Rightarrow x = \frac{3 π}{10} o r x = \frac{7 π}{10}

\sin (x) = \frac{1 - ϕ}{2} \Rightarrow x = \frac{11 π}{10}

x = {\frac{3 π}{10}, \frac{7 π}{10}, \frac{11 π}{10}, \frac{3 π}{2}}

Commented by Tinkutara last updated on 24/Jul/17

Thanks Sir!