Solve the equation algebraically

Question

Solve the equation algebraically, and then use a calculator to find the values on the interval [0,2⁢π). Round your answers to 4 decimal places.

sin2⁡x-2⁢cos2⁡x=0

x=

Answer 1

sin2x -2cos2x = 0

sin2x = 2cos2x

sin2x/cos2x = 2

tan2x = 2

2x = tan^-1(2)

x = (arctan2)/2

= 31.7175 degrees, rounded to 4 decimal places. Use a calculator with inverse trig functions

x = 31.7175 is the quadrant I solution

where tangents are positive

quadrant III also has positive tangents,

so 31.7175+180 = 211.7175 degrees

Answer 2

pass the cos to the other side

sin(2x) = 2cos(2x)

divide by cos(2x)

sin(2x)/cos(2x) = 2

now, we know that sin/cos=tan

tan(2x) = 2

apply the inverse function (arctan) to isolate 2x

2x = arctan(2)

now we use our calculator and we get that

2x = 63.434948

divide by 2

x = 31,717474

round it

x = 31,7175

Answer 3

tan2x = 2; 2x = tan^-12 + nπ; x = 1/2tan^-12 + nπ/2

n = 0; x = 1/2tan^-12 = 0.5536;

n = 1; x = 1/2tan^-12 + π/2 = 2.1244

n =2; x = 1/2tan^-12 + π = 3.6952

n = 3; x = 1/2tan^-12 + 3π/2 = 5.2660