Solve the equation:
\[ 2\cos^2(x) - 5\cos(x) + 2 = 0 \]
Answer (3)
*
The equation given is:
2cos²(x) - 5cos(x) + 2 = 0
To factor this equation, we will use the quadratic formula shown in the attached image.
From the given equation:
a = 2
b = -5
c = 2
This means that:
either cos(x) = 2 ( 2 ) 5 + ( − 5 ) 2 − 4 ( 2 ) ( 2 ) = 2 .......> This solution is rejected as the value of the cosine function lies between -1 and 1 only.
or cos(x) = 2 ( 2 ) 5 − ( − 5 ) 2 − 4 ( 2 ) ( 2 ) = 0.5 ......> This solution is accepted as it lies within -1 and 1
Now, using the inverse of the cosine, we can find that:
x = 3 π + 2 nπ or x = 3 − π + 2 nπ
where "n" is an integer that belongs to Z.
Hope this helps :)
\\\\2t^2-5t+2=0\\\\a=2;\ b=-5;\ c=2\\\\\Delta=b^2-4ac;\ \Delta=(-5)^2-4\cdot2\cdot2=25-16=9\\\\t_1=\frac{-b-\sqrt\Delta}{2a};\ t_2=\frac{-b+\sqrt\Delta}{2a}\\\\t_1=\frac{5-\sqrt9}{2\cdot2}=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}\in < -1;\ 1 >\\\\t_2=\frac{5+\sqrt9}{2\cdot2}=\frac{5+3}{4}=\frac{8}{4}=2\notin < -1;\ 1 >"> 2 co s 2 x − 5 cos + 2 = 0 cos x = t ∈< − 1 ; 1 > 2 t 2 − 5 t + 2 = 0 a = 2 ; b = − 5 ; c = 2 Δ = b 2 − 4 a c ; Δ = ( − 5 ) 2 − 4 ⋅ 2 ⋅ 2 = 25 − 16 = 9 t 1 = 2 a − b − Δ ; t 2 = 2 a − b + Δ t 1 = 2 ⋅ 2 5 − 9 = 4 5 − 3 = 4 2 = 2 1 ∈< − 1 ; 1 > t 2 = 2 ⋅ 2 5 + 9 = 4 5 + 3 = 4 8 = 2 ∈ / < − 1 ; 1 >
cos x = 2 1 → x = 3 π + 2 kπ ∨ x = − 3 π + 2 kπ ( k ∈ Z )
To solve the equation 2 cos 2 ( x ) − 5 cos ( x ) + 2 = 0 , we substitute t = cos ( x ) to form a quadratic equation. The valid solution obtained is cos ( x ) = 2 1 , leading to general solutions of x = 3 π + 2 kπ and x = − 3 π + 2 kπ .
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