The graphical method is fine if you have access to a suitable plotter, if you haven't, having only a pocket calculator, your stuck.
Here's the mathematicians method of solution.
Begin with a substitution using the trig identity
cos2A=2cos2A−1, from which 2cos2A=cos2A+1.
(in order to get the angles on both sides the same).
That gets you
3sin2x=cos2x+2,
which is rewritten as
3sin2x−cos2x=2.
Now consider the trig identity
Rsin(2x−α)=Rsin2xcosα−Rcos2xsinα.
Comparing with the LHS of the equation,
Rcosα=3, and Rsinα=1,
so, squaring and adding, R2=32+12=10, so R=√10,
and dividing, tanα=1/3.
That gets us
√10sin(2x−α)=2,
sin(2x−α)=2/√10,2x−α=sin−1(2/√10),
2x−α=39.23,140.77,399.23,500.77,… deg,
and with alpha = 18.43 deg, that leads to (0 - 360 deg, 2dp),
x=28.83,79.60,208.83,259.60,… deg.
3sin2x=2cos^2x+1
Here's a graphical solution.........https://www.desmos.com/calculator/ynevyjlzji
There are four solutions on [0, 360] degrees......these occur at about 28.8°, 79.6°, 208.8° and 259.6°
The graphical method is fine if you have access to a suitable plotter, if you haven't, having only a pocket calculator, your stuck.
Here's the mathematicians method of solution.
Begin with a substitution using the trig identity
cos2A=2cos2A−1, from which 2cos2A=cos2A+1.
(in order to get the angles on both sides the same).
That gets you
3sin2x=cos2x+2,
which is rewritten as
3sin2x−cos2x=2.
Now consider the trig identity
Rsin(2x−α)=Rsin2xcosα−Rcos2xsinα.
Comparing with the LHS of the equation,
Rcosα=3, and Rsinα=1,
so, squaring and adding, R2=32+12=10, so R=√10,
and dividing, tanα=1/3.
That gets us
√10sin(2x−α)=2,
sin(2x−α)=2/√10,2x−α=sin−1(2/√10),
2x−α=39.23,140.77,399.23,500.77,… deg,
and with alpha = 18.43 deg, that leads to (0 - 360 deg, 2dp),
x=28.83,79.60,208.83,259.60,… deg.