How do you solve: 2sin^2x-sinx-1=0 ?

Question

43566 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-27T23:14:50

#x = pi/2 + 2pin, {7pi}/6 + 2pin, {11pi}/6 + 2pin#

#2sin^2x - sinx - 1 = 0#

Treat the equation as a quadratic equation.

#(2sinx + 1)(sinx - 1) = 0#

#2sinx + 1 = 0#

#sinx = -1/2#

#color(blue)(x = {7pi}/6 + 2pin, {11pi}/6 + 2pin)#

#sinx - 1 = 0#

#sinx = 1#

#color(blue)(x = pi/2 + 2pin)#

Final Answer: #color(green)(x = pi/2 + 2pin, {7pi}/6 + 2pin, {11pi}/6 + 2pin)#

(Where #n# is an integer.)