**Unformatted text preview: **2 x+cos 2 x=1, we can write it as sin 2 x =1-cos 2 x b) Substitute this in our equation 3 cos (x) +2sin 2 (x)=0; 3 cos x +2( 1-cos 2 x )= 0 3 cos x +2 -2cos 2 x = 0 We can write it as -2cos 2 x+3 cos x +2=0 If we suppose cos(x)=y; we can write it as-2y 2 +3y+2=0 c) This expression follows the form ax 2 +bx+c=0, which can be solved using the quadratic formula ¿ − b± √ b 2 − 4 ac 2 a , where a=-2, b=3 and c=2 y = − 3 ± √ 3 2 − 4 ( − 2 ) 2 2 (− 2 ) y = − 3 ± √ 25 − 4 y=-1/2 and y=2 Hence, cos (x)= -1/2 or cos (x)= 2 d) First, we solve cos (x)= -1/2. We do the inverse of the cos of both sides of the equation to extract x inside cos. The inverse of cos (x) is arccos(x) X=arccos (-1/2) X= 2 π 3 = 120° Second, we must solve cos (x)= 2. But there is no solution for this x in R....

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- Fall '08
- BROWNAWELL,WOODRO