System of trigonometric equations:-

Adding Both Equations we Get

10(SinX+ CosX)=10 $\sqrt{2}$ Thus $sinx+ cosx$ = $\sqrt{2}$ -----(1) Squaring (1) Both Sides We Get

$Sin^{2}x$ + $Cos^{2}x$ + $2SinxCosx$ = $2$ Using

$Sin^{2}x$ + $Cos^{2}x$ =1

And $2SinxCosx$ = $Sin2x$

We get 1+ Sin2X= 2 Thus Sin2X=1

Since 0<X<90

And Since Sin90 =1

Therefore Sin2X=Sin90 Thus Equating Both Sides 2X=90 And Thus X= $\boxed{45°}$

With few trigonometry :

Let us call S=sin x ; C=cos X ; we have

{ 4S +2C = 3 $\sqrt{2}$

{ 6S +8C = 7 $\sqrt{2}$

It is a system, 2 equations, with 2 unknown. The determinant = 4x8 - 6x2 is not zero, so there is a single solution (S;C).

If we take 4x second equation less 6x first equation, the S disappear, and we have an equation in C (one unknown, one solution C).

If we take 8x first equation less 2x second, C disappear : one equation in S - one unknown, one solution S.

(We can otherwise, mechanically, use determinant formula, S = $\sqrt{2}$ x $\left| \begin{array}{c}3 3&2\\7&8\end{array}\right|$ / $\left| \begin{array}{c}4 4&2\\6&8\end{array}\right|$ = $\sqrt{2} \times (24-14)$ / $(32-12)$ = $\sqrt{2}$ )

(We can otherwise do other sly method, i personally divided each equation by 2 (to get low coefficient, though making 2nd member become fractional), then subtracted and kept 2 equations among low coefficients ones till elimination of S, and get C= $\frac{\sqrt{2}}{2}$ , then reinjected C value in low coefficient equation to get S.)

Having solved this 2x2 system, you shall get C=S= $\frac{\sqrt{2}}{2}$

The statement tells we shall be in first quadrant. Only x=45° gives sinx = cosx= $\frac{\sqrt{2}}{2}$ (The other solution, with C and S negative, is eliminated by the statement.)

You can use elimination, but you can also do it using this method . He gives an example in this video . However, you still need to consider both equations as each has two solutions in the interval $0<x<90$ . Another possible solution is to divide both sides of the equation by $\cos{x}$ , then square both sides, and then replace $\sec^2{x}$ with $1+\tan^2{x}$ . Then when you solve for $\tan{x}$ in both equations, you will get two rational (and valid) values. Simply pick the one that matches in both equations; then take the arctangent. However, "inspection" works just fine for this one.

Multiplying both equations by 4 & 6 we get ..

24 sinX + 32 cosX = 28 root 2 ..... (1)

24 sinX + 12 cosX = 18 root 2 ..... (2)

(1) - (2)

20 cosX = 10 root 2 => cosX = 1/root 2 =>

X = cos inverse (1/root 2)

so, X = 45