Solve the following trigonometric equations on the interval [0,2pi)?

Arcsin((5 -?33)/4) 2.95 or 6.47 (3s.f.) <=== ANSWER! P.S. My answer is different from the rest of the answerers because I use exact value to find arcsine.

Using The Identity, Sin^2x = 1/2 - 1/2 Cos2x :- Cos 2x = 1 - 2Sin^2x Hence, cos(2x) + 5sin(x) = 1 - 2Sin^2x + 5Sinx = 0 2Sin^2x - 5Sinx - 1 = 0 Using Quadratic Formula -b +- sqrt(b^2-4ac)/2a a = 2 be = -5 c = -1 Sin x = -0.186 (Negative Value) Sin x = 2.686 (Not Possible) Hence, x = Arcsin 0.186 = 0.187 Radians (Ignoring The Negative Sign) Basic Angle = 0.187 Rads Sin Is Negative In The Third & Fourth Quadrants In The Third Quadrant, x = pi + 0.187 = 3.328 Rads In The Fourth Quadrant, x = 2pi - 0.187 = 6.096 Rads 2) 2sinx + 3sinx = 0 5sinx = 0 sinx = 0 x = 0, pi, 2pi Radians.

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