Sunday

December 21, 2014

December 21, 2014

Posted by **mary** on Saturday, February 27, 2010 at 8:55pm.

Find all solutions to the equation in the interval [0,2pi)

cos2x=sinx

I know I have to use some sort of identity, but I have no idea how to go about to solve this.

- Trig. -
**MathMate**, Saturday, February 27, 2010 at 9:25pmYes, by making use of the identity

cos2x = cos²x-sin²x=1-2sin²x, the given equation can be transformed to:

2sin²x+sinx-1=0

which is a quadratic in sin(x).

Solving the quadratic by factoring,

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

or sin(x)=1/2 or sin(x)=-1

Find all roots in the interval [0,2π].

- Trig. -
**mary**, Saturday, February 27, 2010 at 10:14pmThank you very much for the help

- Trig. :) -
**MathMate**, Sunday, February 28, 2010 at 8:16amYou are welcome!

**Answer this Question**

**Related Questions**

Precalculus - Please help!!!!!!!!!!! Find all solutions to the equation in the ...

math - the problem is 2cos^2x + sinx-1=0 the directions are to "use an identity ...

Math - Can I please get some help on these questions: 1. How many solutions does...

Pre-Calculus - Find all solutions to the equation in the interval [0, 2pi) cos4x...

calculus - Find all solutions to the equation in the interval [0,2pi) Cosx-cos2x

identities trig? - find all solutions to the equation in the interval [0,2pie] ...

trig - Find the solutions to the equation cos2x=sinx and 2cos^2x-3cosx-3=0

Trig - Solve Sinx=Cos2x-1 for all values between 0 and 2pi

Trig - Find all solutions of the equation on the interval [0,2pi): Tan^2x=1-secx

Trig - Find all solutions of equation on interval [0,2pi] 1=cot^2x + cscx