Thread: finding solutions

PLEASE help finding solutions on the [0,2PI)


1.
cos[Pi/2 + X] + sin(x-PI) = 1

2.
2sinxcosx+cosx=0

3.
cos2x+sinx=1

4.
cosx=sinx

5.
2sin^2x=sinx

6.
tanx+secx=1


please I am so confused. Thanks.

Have you studied algebra or trigonometry at all? If you've studied algebra, have you done any equation-solving? If you've studied trig, have you learned anything about trigonometric identities?

Originally Posted by lisasmith234

1. cos[Pi/2 + X] + sin(x-PI) = 1

This requires some trig identities to get started (angle-sum, angle-difference, or "relating the sine and cosine" rules).

Originally Posted by lisasmith234

2. 2sinxcosx+cosx=0

This factors (which they'll cover in algebra) as cos(x)[(2 sin(x) + 1] = 0. Then you'd solve each factor, resulting in the trig equations cos(x) = 0 and sin(x) = -1/2. Then you'd use memorized "reference angle values" to find the values of x.

Originally Posted by lisasmith234

3. cos2x+sinx=1

This one would use one of a couple trig identities to restate this only in terms of the sine: 1 - 2 sin^2(x) + sin(x) = 1. Then you'd solve the quadratic equation (using factoring or the Quadratic Formula), and then solve the two resulting trig equations in the same way as in exercise (2).

Originally Posted by lisasmith234

4. cosx=sinx

A good start would be to divide through by the cosine. This would give you tan(x) = 1, which you'd solve using memorized values.

Originally Posted by lisasmith234

5. 2sin^2x=sinx

A good start would be to move everything over to one side of the "equals" sign, and then solve the resulting quadratic, similar to exercises (2) and (3).

Originally Posted by lisasmith234

6. tanx+secx=1

One method might be to restate the tangent and the secant in terms of sines and cosines, and then go from there.

Please reply with specifications of your needs, so that we can help you get "unconfused" on the algebra and trig involved here. Thank you!