What is the general solution for 2cos^2(x)+3sin(x)-3=0,0<= x<= 360 ?

The general solution for 2cos^2(x)+3sin(x)-3=0,0<= x<= 360 is x=30,x=150,x=90

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2cos^2(x)+3sin(x)-3=0,0<= x<= 360

2 cos^{2} (x) + 3 sin (x) - 3 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 3 0^{\circ}, x = 15 0^{\circ}, x = 9 0^{\circ}

Solution steps

2 cos^{2} (x) + 3 sin (x) - 3 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = 1

sin (x) = \frac{1}{2}, 0 \leq x \leq 36 0^{\circ} : x = 3 0^{\circ}, x = 15 0^{\circ}

sin (x) = 1, 0 \leq x \leq 36 0^{\circ} : x = 9 0^{\circ}

Combine all the solutions

x = 3 0^{\circ}, x = 15 0^{\circ}, x = 9 0^{\circ}

5 cot (θ) + 7 = 0

cos (x) = - 4

- 2 cos (b) + 4 = cos (b) + 6

25 csc^{2} (θ) - 4 = 0

tan (θ) = \frac{1}{5}

What is the general solution for 2cos^2(x)+3sin(x)-3=0,0<= x<= 360 ?
The general solution for 2cos^2(x)+3sin(x)-3=0,0<= x<= 360 is x=30,x=150,x=90