20-02-2021
Mathematics

contestada

solve for x 0°<x<360
3cos2x +sinx=1

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Аноним Аноним

20-02-2021

Step-by-step explanation:

3cos2x +sinx=1

3(1-2sin²x)+sinx=1

3-6sin²x+sinx=1

sinx(-6sinx+1)=1

either

sinx=1

sinx=sin(90)

:.x=90

or

-6sinx+1=1

-6sinx=1-1

sinx=0/-6

sinx=sin0,sin180

:.x=0,180and90

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