sin^2 x+cos^2 (2x)+sin^2 (3x)=(3/2)


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Question Number 203325 by depressiveshrek last updated on 16/Jan/24

$\sin^{2} x + \cos^{2} (2 x) + \sin^{2} (3 x) = \frac{3}{2}$
	Answered by esmaeil last updated on 16/Jan/24
	$sin^2 x+(1−2sin^2 x)^2 + (3sinx−4sin^3 x)^2 =(3/2)→ sin^2 x+1+4sin^4 x−4sin^2 x+ 9sin^2 x+16sin^6 x−24sin^4 x=(3/2)→ 16sin^6 x−20sin^4 x+6sin^2 x=(3/2)→ 32m^6 −40m^4 +12m^2 −3=0→ 4(2m^2 )^3 −5(2m^2 )^2 +6(2m^2 )−3=0 4p^3 −5p^2 +6p−3=0 4p^3 −4p^2 +4p−p^2 +2p−4+1=0 4(p^3 −p^2 +p)−(p−1)^2 −4=0 4(p^3 −p^2 +p−1)−(p−1)^2 =0 4(p−1)(p^2 +1)−(p−1)^2 =0 (p−1)(4p^2 +4−p+1)=0 p=1→2m^2 =1→m=±((√2)/2)→ sinx=sin(±(π/4))→ { ((x=2kπ±(π/4))),((x=2kπ+((3π)/4),x=2kπ+((5π)/4))) :} 4p^2 −p+5=0×$
	$Missing \left or extra \right$ ${(3 s i n x - 4 {s i n}^{3} x)}^{2} = \frac{3}{2} \to$ ${s i n}^{2} x + 1 + 4 {s i n}^{4} x - 4 {s i n}^{2} x +$ $9 {s i n}^{2} x + 16 {s i n}^{6} x - 24 {s i n}^{4} x = \frac{3}{2} \to$ $16 {s i n}^{6} x - 20 {s i n}^{4} x + 6 {s i n}^{2} x = \frac{3}{2} \to$ $32 m^{6} - 40 m^{4} + 12 m^{2} - 3 = 0 \to$ $4 {(2 m^{2})}^{3} - 5 {(2 m^{2})}^{2} + 6 (2 m^{2}) - 3 = 0$ $4 p^{3} - 5 p^{2} + 6 p - 3 = 0$ $4 p^{3} - 4 p^{2} + 4 p - p^{2} + 2 p - 4 + 1 = 0$ $4 (p^{3} - p^{2} + p) - {(p - 1)}^{2} - 4 = 0$ $4 (p^{3} - p^{2} + p - 1) - {(p - 1)}^{2} = 0$ $4 (p - 1) (p^{2} + 1) - {(p - 1)}^{2} = 0$ $(p - 1) (4 p^{2} + 4 - p + 1) = 0$ $p = 1 \to 2 m^{2} = 1 \to m = \pm \frac{\sqrt{2}}{2} \to$ $s i n x = s i n (\pm \frac{π}{4}) \to {\begin{cases} x = 2 k π \pm \frac{π}{4} \\ x = 2 k π + \frac{3 π}{4}, x = 2 k π + \frac{5 π}{4} \end{cases}$ $4 p^{2} - p + 5 = 0 \times$
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