Trigonometry Question

Question

Solve for x in the given interval.

2^1/2sinx +tan x=0, -180 degree < x < 0 degree

Answer 1

sqr2(sinx) + tanx = 0

sqr2(sinx) + sinx/cosx = 0

sinx(sqr2 + 1/cosx) = 0 set each factor = 0 and solve for x

sinx = 0

x = -180 or 0 degrees, but there is no value of x such that sinx=0 in the open interval (-180,0)

-180<x<0 means x can't be the endpoints

the only x solution would be the x that satisfies sqr2 + 1/cosx = 0

1/cosx =-sqr2

cosx = -1/sqr2

That's an x in quadrant III or II where cosines are negative

-180 < x < -90 in quadrant III

x = -135 degrees

reference angle is 45. cos45 = sqr2/2 = 1/sqr2

add -180 + 45 = -135 degrees