What is the general solution for 2cos^2(x)-7cos(x)=-3,0<= x<= 2pi ?

The general solution for 2cos^2(x)-7cos(x)=-3,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3

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2cos^2(x)-7cos(x)=-3,0<= x<= 2pi

2 cos^{2} (x) - 7 cos (x) = - 3, 0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

2 cos^{2} (x) - 7 cos (x) = - 3, 0 \leq x \leq 2 π

Solve by substitution

cos (x) = 3, cos (x) = \frac{1}{2}

cos (x) = 3, 0 \leq x \leq 2 π :

No Solution

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}

sin (2 x) = \frac{( 4 m - 9 )}{8}

sin (2 x) = - 2

sin (x) = sin (9 0^{\circ} - x)

2 sin (2 θ) - 3 - \frac{6}{sin ( 2 θ ) - 1} = 0

8 cos (B) + 52 = 8

What is the general solution for 2cos^2(x)-7cos(x)=-3,0<= x<= 2pi ?
The general solution for 2cos^2(x)-7cos(x)=-3,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3