Without using a calculator, simplify (sinX + 1) = 0.?

Question

And if you have a hankerin' for trig, solving cosX + 1/2 = 0 without a calculator would be much appreciated. Showing how you did it would also be extremely helpful.

Answer 1

(sinX + 1) = 0---> sinX = -1--->x = 3pi/2 + (2pi)n, radians, or x = 270 + 360n degrees, where n is any whole number including 0.

The first step was simple algebra. In the second step, I just remembered that the sine is how far up on the unit circle that angle points, so -1 means it points straight down, which is 3/4 of a circle. Adding a whole circle puts it right back where it started from.

cosX + 1/2 = 0---> cosX = -1/2 ---> x = 2pi/3 or 4pi/3 + (2pi)n radians, 120 or 240 + 360n degrees.

Again, the first step was simple algebra. As for the second, my trig teacher made us memorize the sines/cosines of 30, 45, and 60 degrees, so I just know that the cosine of 60 degrees is 1/2. Go around to the other side of the unit circle and the cosine of 120 degrees is -1/2. Go down to the bottom of the unit circle, and the cosine of 240 degrees is also -1/2.

Answer 2

sinX+1=0 then sinX=-1 (minus 1) so X=270 degrees


cosX+1/2=0 Then ,cosX=-1/2(minus 1/2) so X=240 degrees

Answer 3

sinx + 1 = 0
sinx = -1
x = 270, since sin is -1 at that angle

cosx + 1/2 = 0
cosx = -1/2
x = 120, 240 since cosx = 1/2 and in the 2nd and 3rd quadrants cosine is negative

No, i didnt use a calculator to find the values, its simply a matter of knowing your trigonometric ratios and whether theyre positive or negative. In the first quadrant, everything is positive. In the second quandrant, the sine function and the cosecant function are positive, and all other functinos are negative. in the 3rd quadrant, the tangent and cotangent are positive, the rest negative. And finally, in the 4th quadrant, the cosine and the secant functions are positive, the rest negative.

Answer 4

Simplifying
sin x + 1 = 0
sin x = -1

Solving it:
x = arcsin(-1) = -(pi)/2 + 2(pi)n, n is any integer

Or with the restricted domain of [0,2(pi)], also [0,360]

x = 3(pi)/2 or 270 degrees

---

cos x + 1/2 = 0
cos x = -1/2
x = arccos(-1/2)

From a trig sheet or by drawing a unit circle,

x = arccos(-1/2) = 2(pi)/3 + 2(pi)n
and
x = 4(pi)/3 + 2(pi)n

Or with restricted domain of [0,2(pi)], also [0,360]

x = 2(pi)/3 radians or 120 degrees
and
x = 4(pi)/3 radians or 240 degrees

Sorry for the interchanging of radians and degrees

Answer 5

sinX=-1.