+118702 There are a number of different ways to do a question like this. Here are some ideas.
http://www.thestudentroom.co.uk/showthread.php?t=1586679
ok now I will give it a go.
\begin{array}{rlll} cos^2\theta+sin^2\theta&=&1\\ cos\theta&=&\sqrt{1-sin^2\theta\\\\ 3cosx-4sinx &=&1\\ 3\sqrt{1-sin^2x}-4sinx &=&1\qquad &\mbox{ }\\ 3\sqrt{1-sin^2x} &=&1+4sinx\qquad &\mbox{Square both sides }\\ 9\times(1-sin^2x) &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 9-9sin^2x &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 0 &=&-8+8sinx+25sin^2x\qquad &\mbox{}\\ 25sin^2x+8sinx-8 &=&0\qquad &\mbox{}\\ Let \;y=sinx&\\ 25y^2+8y-8 &=&0\qquad &\mbox{}\\ y&=&\frac{-8\pm \sqrt{64+800}}{50}\\ sinx&=&\frac{-8\pm \sqrt{16*9*6}}{50}\\ sinx&=&\frac{-8\pm 12\sqrt{6}}{50}\\ \end{array}
sin360∘−1((−8−√864)50)=−48.40685667866∘
sin360∘−1((−8+√864)50)=25.332938613029∘
These answers should be checked by substituting them back into the original equation but I am going to leave that to you. (I'm too tired) 
+118702 There are a number of different ways to do a question like this. Here are some ideas.
http://www.thestudentroom.co.uk/showthread.php?t=1586679
ok now I will give it a go.
\begin{array}{rlll} cos^2\theta+sin^2\theta&=&1\\ cos\theta&=&\sqrt{1-sin^2\theta\\\\ 3cosx-4sinx &=&1\\ 3\sqrt{1-sin^2x}-4sinx &=&1\qquad &\mbox{ }\\ 3\sqrt{1-sin^2x} &=&1+4sinx\qquad &\mbox{Square both sides }\\ 9\times(1-sin^2x) &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 9-9sin^2x &=&1+8sinx+16sin^2x\qquad &\mbox{}\\ 0 &=&-8+8sinx+25sin^2x\qquad &\mbox{}\\ 25sin^2x+8sinx-8 &=&0\qquad &\mbox{}\\ Let \;y=sinx&\\ 25y^2+8y-8 &=&0\qquad &\mbox{}\\ y&=&\frac{-8\pm \sqrt{64+800}}{50}\\ sinx&=&\frac{-8\pm \sqrt{16*9*6}}{50}\\ sinx&=&\frac{-8\pm 12\sqrt{6}}{50}\\ \end{array}
sin360∘−1((−8−√864)50)=−48.40685667866∘
sin360∘−1((−8+√864)50)=25.332938613029∘
These answers should be checked by substituting them back into the original equation but I am going to leave that to you. (I'm too tired)
