### Video Transcript

Solve root two sin 𝜃 plus root
three cos 𝜃 equals two, where 𝜃 is greater than zero and less than or equal to two
𝜋. Give your answer in radians to two
decimal places.

In this question, we have a
harmonic equation that we can solve by firstly recalling one of the addition rules
of trigonometry. This states that sin of 𝐴 plus 𝐵
is equal to sin 𝐴 cos 𝐵 plus cos 𝐴 sin 𝐵. We begin by writing any harmonic
equation in the form 𝑎 sin 𝜃 plus 𝑏 cos 𝜃 as 𝑅 multiplied by sin of 𝜃 plus
𝛼. In this question, we want to
rewrite root two sin 𝜃 plus root three cos 𝜃 in the form 𝑅 multiplied by sin of
𝜃 plus 𝛼, where the values of 𝑅 and 𝛼 can be determined using the constants 𝑎
and 𝑏.

We may simply wish to recall that
𝑅 is equal to the square root of 𝑎 squared plus 𝑏 squared and 𝛼 is equal to the
inverse tan of 𝑏 over 𝑎. Alternatively, we can rewrite the
right-hand side of our equation using the addition formula. After distributing the parentheses
or expanding the brackets, we have 𝑅 sin 𝜃 cos 𝛼 plus 𝑅 cos 𝜃 sin 𝛼. We can then compare the
coefficients of sin 𝜃 and cos 𝜃 on each side of the equation, giving us root two
is equal to 𝑅 cos 𝛼 and root three is equal to 𝑅 sin 𝛼.

We now have a pair of simultaneous
equations that we can use to find the values of 𝛼 and 𝑅. Dividing equation two by equation
one, we have root three over root two is equal to 𝑅 sin 𝛼 over 𝑅 cos 𝛼. Since sin 𝛼 over cos 𝛼 is equal
to tan 𝛼, we have tan 𝛼 is equal to root three over root two. We can then take the inverse
tangent of both sides of this equation such that 𝛼 is equal to the inverse tan of
root three over root two. We can therefore see that 𝛼 is
equal to the inverse tan of 𝑏 over 𝑎. Typing this into our calculator,
ensuring it is in radian mode, we have 𝛼 is equal to 0.886 and so on.

We can then substitute this exact
value of 𝛼 back into equation one or two to calculate 𝑅. In equation two, we have 𝑅 is
equal to root three over sin 𝛼, which is equal to root five. We could also have found this value
of 𝑅 using our knowledge of the Pythagorean trigonometric identity, which leads us
to the fact that 𝑅 is equal to the square root of 𝑎 squared plus 𝑏 squared. The square root of root two squared
plus root three squared is equal to root five. Clearing some space, we now have
the equation root five multiplied by sin of 𝜃 plus 0.886 and so on is equal to
two. And we need to solve this equation
for values of 𝜃 greater than zero and less than or equal to two 𝜋.

We begin by dividing both sides of
our equation by root five. This gives us the sin of 𝜃 plus
0.886 and so on is equal to two over root five. Next, we take the inverse sine of
both sides. 𝜃 plus 0.886 is equal to the
inverse sin of two over root five. Ensuring that our calculator is in
radian mode once again, the right-hand side becomes 1.107 and so on. We can now find one solution to our
equation by subtracting 0.886 and so on from both sides. This gives us 0.221 and so on. Rounding this to two decimal places
as required, we have 𝜃 is equal to 0.22.

At this stage, we might think we
are finished. However, we want all the values of
𝜃 that lie between zero and two 𝜋. One way of checking for any other
solutions is using our CAST diagram. The positive angles between zero
and two 𝜋 are measured in a counterclockwise direction as shown. Since the sine of our angle is
positive, we know there will be solutions in the first and second quadrants. And from the symmetry of the sine
function, we have a second solution where 𝜃 plus 0.886 and so on is equal to 𝜋
minus 1.107 and so on. We can then subtract 0.886 from
both sides, giving us 𝜃 is equal to 1.148 and so on. Once again, rounding to two decimal
places, we have 𝜃 is equal to 1.15.

The solutions to the equation root
two sin 𝜃 plus root three cos 𝜃 equals two, where 𝜃 is greater than zero and less
than or equal to two 𝜋 radians, to two decimal places, are 𝜃 is equal to 0.22 and
1.15.